Similarity of Triangles

Congruence versus Similarity is about comparing shapes in geometry.

Two figures are congruent if they have the same shape and the same size. If you could slide, turn, or flip one figure and make it fit exactly on top of the other, the figures are congruent.

Two figures are similar if they have the same shape, but they may be different sizes. Similar figures look alike, and one is a scaled copy of the other.

Understanding the difference between congruence and similarity is very important in triangle geometry. It helps us decide whether corresponding sides must be equal or only proportional, and whether figures match exactly or just have the same shape.

1. What does congruent mean?

Congruent figures have all matching parts equal.

• Corresponding sides are equal in length.
• Corresponding angles are equal in measure.
• The figures are exactly the same size and shape.

If triangle \(ABC\) is congruent to triangle \(DEF\), we write:

\(\triangle ABC \cong \triangle DEF\)

This means the order matters:

• \(A \leftrightarrow D\)
• \(B \leftrightarrow E\)
• \(C \leftrightarrow F\)

So the corresponding parts satisfy:

$$AB = DE, \quad BC = EF, \quad AC = DF$$

and

$$\angle A = \angle D, \quad \angle B = \angle E, \quad \angle C = \angle F$$

2. What does similar mean?

Similar figures have the same shape, but not necessarily the same size.

• Corresponding angles are equal.
• Corresponding side lengths are proportional.

If triangle \(ABC\) is similar to triangle \(DEF\), we write:

\(\triangle ABC \sim \triangle DEF\)

Then:

$$\angle A = \angle D, \quad \angle B = \angle E, \quad \angle C = \angle F$$

and the side lengths have the same ratio:

$$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$

This common ratio is called the scale factor.

3. The key difference

The most important difference is this:

• Congruent figures: same shape and same size
• Similar figures: same shape, but size can be different

Another way to say this is:

• Congruent figures have corresponding sides with ratio \(1:1\).
• Similar figures have corresponding sides with a constant ratio, which may or may not be \(1:1\).

So every pair of congruent figures is also similar, because equal side lengths are proportional with scale factor \(1\).

But not every pair of similar figures is congruent, because similar figures can have different sizes.

4. How to check for congruence or similarity

When comparing two triangles, ask these questions:

1. Are all matching angles equal?
2. Are the matching sides exactly equal, or only proportional?

If the angles match and the sides are equal, the triangles are congruent.

If the angles match and the sides are proportional, the triangles are similar.

5. Corresponding parts must match correctly

It is very important to compare the correct sides and angles. The order of the letters tells you which parts correspond.

For example, if \(\triangle PQR \sim \triangle XYZ\), then:

• \(P \leftrightarrow X\)
• \(Q \leftrightarrow Y\)
• \(R \leftrightarrow Z\)

So:

$$\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ}$$

If you mix up the matching sides, your ratio will be wrong.

Worked Example 1: Deciding whether figures are congruent or similar

Triangle \(A\) has side lengths \(3\), \(4\), and \(5\).

Triangle \(B\) has side lengths \(3\), \(4\), and \(5\).

Since all three corresponding sides are equal, the triangles are the same size and shape.

Conclusion: The triangles are congruent.

They are also similar, because congruent figures are always similar.

Now compare Triangle \(C\) with side lengths \(6\), \(8\), and \(10\) to Triangle \(A\).

Check the ratios:

$$\frac{6}{3} = 2, \quad \frac{8}{4} = 2, \quad \frac{10}{5} = 2$$

All corresponding sides have the same ratio. So the triangles have the same shape, but Triangle \(C\) is larger.

Conclusion: The triangles are similar, not congruent.

Worked Example 2: Using angle and side information

Suppose two triangles have angle measures:

Triangle 1: \(50^\circ, 60^\circ, 70^\circ\)

Triangle 2: \(50^\circ, 60^\circ, 70^\circ\)

The corresponding angles are equal, so the triangles have the same shape.

Now suppose Triangle 1 has sides \(5, 6, 7\), and Triangle 2 has sides \(10, 12, 14\).

Check the ratios:

$$\frac{10}{5} = 2, \quad \frac{12}{6} = 2, \quad \frac{14}{7} = 2$$

The side lengths are proportional, so the triangles are similar.

They are not congruent because the side lengths are not equal.

Worked Example 3: Finding a missing side in similar triangles

Suppose \(\triangle ABC \sim \triangle DEF\).

Let:

• \(AB = 4\)
• \(BC = 6\)
• \(DE = 10\)
• \(EF = x\)

Because \(\triangle ABC \sim \triangle DEF\), corresponding sides are proportional. From the order, \(AB \leftrightarrow DE\) and \(BC \leftrightarrow EF\).

So:

$$\frac{AB}{DE} = \frac{BC}{EF}$$

Substitute the values:

$$\frac{4}{10} = \frac{6}{x}$$

Cross multiply:

$$4x = 60$$

$$x = 15$$

Answer: \(EF = 15\).

This example shows how similarity lets us find missing lengths using equal ratios.

Worked Example 4: Similar or neither?

Triangle \(MNO\) has sides \(4, 5, 6\).

Triangle \(RST\) has sides \(8, 10, 13\).

Check whether the side lengths are proportional:

$$\frac{8}{4} = 2, \quad \frac{10}{5} = 2, \quad \frac{13}{6} \neq 2$$

Because the ratios are not all equal, the triangles are not similar.

They are also not congruent, because their side lengths are not equal.

Conclusion: These triangles are neither congruent nor similar.

6. Important facts to remember about triangles

• If two triangles are congruent, all corresponding sides and angles are equal.
• If two triangles are similar, all corresponding angles are equal and corresponding sides are proportional.
• Congruent triangles are a special case of similar triangles with scale factor \(1\).
• Similar triangles can be enlarged or reduced versions of each other.

7. Why this matters in geometry

Similarity is used to find missing side lengths, compare shapes, and build geometric proofs.

In triangle geometry, similar triangles help us discover relationships between lengths and angles. Later, this idea is used in important results such as proofs of the Pythagorean theorem.

Congruence is used when we need exact equality. If a problem says two triangles are congruent, then matching sides are equal, not just proportional.

8. Common mistakes

• Thinking similar means exactly equal in size. It does not. Similar means same shape.
• Forgetting that congruent figures are also similar.
• Using the wrong corresponding sides in a ratio.
• Checking only one pair of sides instead of all corresponding sides.
• Assuming equal angles alone mean congruent. Equal angles show same shape, not same size.

Brief Summary

Congruent figures have the same shape and the same size, so all corresponding sides and angles are equal.

Similar figures have the same shape, so corresponding angles are equal and corresponding sides are proportional, but the sizes can be different.

When comparing triangles, always match corresponding parts carefully. If side lengths are equal, the triangles may be congruent. If side lengths have a constant ratio, the triangles may be similar.

Lesson: The Basic Proportionality Theorem (Thales Theorem)

In geometry, triangles often contain smaller shapes inside them. When a line is drawn parallel to one side of a triangle, it creates smaller triangles that are closely related to the original triangle. The Basic Proportionality Theorem, also called Thales Theorem, tells us exactly how the sides are divided in this situation.

This theorem is an important part of the topic Similarity of Triangles. It helps us compare lengths, find unknown sides, and understand why certain ratios in a triangle stay equal.

Statement of the Basic Proportionality Theorem

Consider a triangle \(\triangle ABC\). Let a line through points \(D\) and \(E\) cut sides \(AB\) and \(AC\) respectively, such that \(DE \parallel BC\).

Then the theorem states that the line divides the two sides in the same ratio:

$$ \frac{AD}{DB} = \frac{AE}{EC} $$

So, if a line is parallel to one side of a triangle, it divides the other two sides proportionally.

Visual idea

Imagine triangle \(ABC\). A smaller triangle \(ADE\) is formed inside it, with \(DE\) parallel to the base \(BC\). Since the line is parallel, the small triangle and the large triangle have the same shape. That is why their corresponding sides are in proportion.

Why this theorem is true

Since \(DE \parallel BC\), corresponding angles are equal:

• \(\angle ADE = \angle ABC\)
• \(\angle AED = \angle ACB\)
• \(\angle A\) is common to both triangles

So, triangles \(\triangle ADE\) and \(\triangle ABC\) are similar.

From similarity, corresponding sides are proportional:

$$ \frac{AD}{AB} = \frac{AE}{AC} $$

Now write the whole sides as:

$$ AB = AD + DB, \quad AC = AE + EC $$

Substituting these gives:

$$ \frac{AD}{AD+DB} = \frac{AE}{AE+EC} $$

On simplifying, we get:

$$ \frac{AD}{DB} = \frac{AE}{EC} $$

This proves the Basic Proportionality Theorem.

Important points to remember

• The line must be parallel to one side of the triangle.
• The theorem applies to the other two sides of the triangle.
• The ratios must compare matching divided parts, such as \(AD:DB\) and \(AE:EC\).
• Do not compare incorrect sides, such as \(AD:AE\), unless the question specifically asks for it using similarity.

Different forms of the theorem

If \(DE \parallel BC\) in \(\triangle ABC\), then all of the following are useful:

$$ \frac{AD}{DB} = \frac{AE}{EC} $$ $$ \frac{AD}{AB} = \frac{AE}{AC} $$ $$ \frac{DE}{BC} = \frac{AD}{AB} = \frac{AE}{AC} $$

These come from the similarity of triangles \(\triangle ADE\) and \(\triangle ABC\).

Worked Example 1: Finding one unknown segment

In \(\triangle ABC\), \(D\) lies on \(AB\) and \(E\) lies on \(AC\). Suppose \(DE \parallel BC\), \(AD = 4\) cm, \(DB = 6\) cm, and \(AE = 5\) cm. Find \(EC\).

Using the Basic Proportionality Theorem:

$$ \frac{AD}{DB} = \frac{AE}{EC} $$

Substitute the given values:

$$ \frac{4}{6} = \frac{5}{EC} $$

Simplify:

$$ \frac{2}{3} = \frac{5}{EC} $$

Cross-multiply:

$$ 2 \cdot EC = 3 \cdot 5 $$ $$ 2EC = 15 $$ $$ EC = 7.5 \text{ cm} $$

Answer: \(EC = 7.5\) cm.

Worked Example 2: Finding a part when whole side is known

In \(\triangle ABC\), \(DE \parallel BC\). Point \(D\) divides \(AB\) into \(AD = 3\) cm and \(DB = 9\) cm. If \(AC = 16\) cm, find \(AE\) and \(EC\).

First use similarity in the form:

$$ \frac{AD}{AB} = \frac{AE}{AC} $$

Find \(AB\):

$$ AB = AD + DB = 3 + 9 = 12 \text{ cm} $$

Now substitute:

$$ \frac{3}{12} = \frac{AE}{16} $$ $$ \frac{1}{4} = \frac{AE}{16} $$

So,

$$ AE = 16 \times \frac{1}{4} = 4 \text{ cm} $$

Now find \(EC\):

$$ EC = AC - AE = 16 - 4 = 12 \text{ cm} $$

Answer: \(AE = 4\) cm and \(EC = 12\) cm.

Worked Example 3: Finding a missing segment algebraically

In \(\triangle ABC\), \(DE \parallel BC\). Let \(AD = x\), \(DB = 4\), \(AE = 6\), and \(EC = 8\). Find \(x\).

Use the theorem:

$$ \frac{AD}{DB} = \frac{AE}{EC} $$

Substitute:

$$ \frac{x}{4} = \frac{6}{8} $$

Simplify the right side:

$$ \frac{x}{4} = \frac{3}{4} $$

Therefore,

$$ x = 3 $$

Answer: \(AD = 3\).

Worked Example 4: Using the theorem to check whether a line is parallel

In \(\triangle ABC\), point \(D\) lies on \(AB\) and point \(E\) lies on \(AC\). Suppose:

$$ AD = 2, \quad DB = 3, \quad AE = 4, \quad EC = 6 $$

We want to check whether \(DE \parallel BC\).

Compare the ratios:

$$ \frac{AD}{DB} = \frac{2}{3} $$ $$ \frac{AE}{EC} = \frac{4}{6} = \frac{2}{3} $$

Since

$$ \frac{AD}{DB} = \frac{AE}{EC} $$

the two sides are divided in the same ratio. So, by the converse idea, the line \(DE\) is parallel to \(BC\).

Answer: Yes, \(DE \parallel BC\).

How to solve questions on this theorem

1. Identify the triangle and the line parallel to one side.
2. Write the correct proportionality relation.
3. Substitute the known lengths carefully.
4. Solve the equation.
5. Check whether the answer makes sense in the figure.

Common mistakes to avoid

• Forgetting the parallel condition: The theorem works only when the line is parallel to one side.
• Using the wrong segments: If you use \(AD\), then compare it with \(DB\), not with the full side unless you are using a different correct form.
• Adding sides incorrectly: Remember that \(AB = AD + DB\) and \(AC = AE + EC\).
• Mixing corresponding sides: Keep the order of ratios consistent.

Quick practice questions

• If \(AD = 5\), \(DB = 10\), and \(AE = 6\), find \(EC\) when \(DE \parallel BC\).
• If \(AD = 4\), \(AB = 10\), and \(AC = 15\), find \(AE\).
• If \(AD = 3\), \(DB = 7\), \(AE = 6\), and \(EC = 14\), check whether \(DE \parallel BC\).

Brief Summary

The Basic Proportionality Theorem says that if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

$$ \frac{AD}{DB} = \frac{AE}{EC} $$

This theorem comes from the similarity of triangles. It is very useful for finding unknown lengths and proving relationships in geometry.

Converse of the Basic Proportionality Theorem

In triangle geometry, a very useful idea is that a line parallel to one side of a triangle divides the other two sides in the same ratio. This is called the Basic Proportionality Theorem (BPT), also known as Thales' theorem.

But sometimes, questions work the other way around. Instead of telling us that a line is parallel, the question gives us proportional lengths on two sides of a triangle and asks us to decide whether the line is parallel to the third side. That is where the Converse of the Basic Proportionality Theorem is used.

This lesson will help you understand what the converse says, when to use it, how to check the ratios correctly, and how to solve typical exam-style questions.

1. Recall: Basic Proportionality Theorem

In a triangle, if a line is drawn parallel to one side to intersect the other two sides, then it divides those two sides in the same ratio.

For example, in triangle \(ABC\), if a line through points \(D\) on \(AB\) and \(E\) on \(AC\) is such that \(DE \parallel BC\), then

$$\frac{AD}{DB}=\frac{AE}{EC}$$

This is the original theorem.

2. What does “converse” mean?

The converse of a statement means reversing it.

• Original theorem: If a line is parallel to one side of a triangle, then it divides the other two sides proportionally.
• Converse: If a line divides two sides of a triangle proportionally, then the line is parallel to the third side.

3. Statement of the Converse of the Basic Proportionality Theorem

In triangle \(ABC\), let \(D\) be a point on \(AB\) and \(E\) be a point on \(AC\). If

$$\frac{AD}{DB}=\frac{AE}{EC},$$

then

$$DE \parallel BC$$

This means that if the two sides are cut in the same ratio, the joining line must be parallel to the third side.

4. Visual idea

Imagine triangle \(ABC\). Point \(D\) lies somewhere on side \(AB\), and point \(E\) lies somewhere on side \(AC\). Now join \(D\) and \(E\).

If the pieces on side \(AB\) and the pieces on side \(AC\) satisfy

$$\frac{AD}{DB}=\frac{AE}{EC},$$

then the line segment \(DE\) is forced to be parallel to \(BC\).

So this theorem is a test for parallel lines inside a triangle.

5. Why this theorem matters

This theorem helps us:

• prove that two lines are parallel,
• check whether a given figure is correctly drawn,
• solve missing-length problems,
• use similarity of triangles more effectively.

6. Important condition to remember

The theorem works only when:

• \(D\) lies on one side of the triangle,
• \(E\) lies on another side of the same triangle,
• the two sides are divided proportionally.

You must compare the matching parts carefully. Usually, we compare:

$$\frac{AD}{DB} \text{ and } \frac{AE}{EC}$$

or sometimes the equivalent form

$$\frac{AD}{AB}=\frac{AE}{AC}$$

Both forms can be useful, but only if the corresponding parts are matched correctly.

7. How to use the theorem step by step

1. Identify the triangle.
2. Find the two points on two sides of the triangle.
3. Write the ratios of the divided parts.
4. Check whether the ratios are equal.
5. If they are equal, conclude that the joining line is parallel to the third side.

8. Worked Example 1: Direct ratio check

In triangle \(ABC\), point \(D\) lies on \(AB\) and point \(E\) lies on \(AC\). Given:

• \(AD = 3\) cm, \(DB = 6\) cm
• \(AE = 2\) cm, \(EC = 4\) cm

Show that \(DE \parallel BC\).

Solution:

First, write the two ratios:

$$\frac{AD}{DB}=\frac{3}{6}=\frac{1}{2}$$ $$\frac{AE}{EC}=\frac{2}{4}=\frac{1}{2}$$

Since

$$\frac{AD}{DB}=\frac{AE}{EC},$$

the two sides are divided in the same ratio.

Therefore, by the Converse of the Basic Proportionality Theorem,

$$DE \parallel BC$$

9. Worked Example 2: Ratios not equal

In triangle \(PQR\), point \(S\) lies on \(PQ\) and point \(T\) lies on \(PR\). Given:

• \(PS = 4\) cm, \(SQ = 5\) cm
• \(PT = 6\) cm, \(TR = 9\) cm

Is \(ST \parallel QR\)?

Solution:

Check the ratios:

$$\frac{PS}{SQ}=\frac{4}{5}$$ $$\frac{PT}{TR}=\frac{6}{9}=\frac{2}{3}$$

Now compare:

$$\frac{4}{5} \ne \frac{2}{3}$$

The ratios are not equal.

So the two sides are not divided proportionally.

Therefore, by the converse of BPT, we cannot say that

$$ST \parallel QR$$

In fact, \(ST\) is not parallel to \(QR\).

10. Worked Example 3: Using the theorem to find an unknown

In triangle \(ABC\), point \(D\) lies on \(AB\) and point \(E\) lies on \(AC\). Suppose:

• \(AD = 4\) cm
• \(DB = 8\) cm
• \(AE = 5\) cm
• \(EC = x\) cm

If \(DE \parallel BC\), find \(x\).

Solution:

Since \(DE \parallel BC\), the Basic Proportionality Theorem gives:

$$\frac{AD}{DB}=\frac{AE}{EC}$$

Substitute the values:

$$\frac{4}{8}=\frac{5}{x}$$ $$\frac{1}{2}=\frac{5}{x}$$

Cross-multiply:

$$x = 10$$

So,

$$EC = 10 \text{ cm}$$

Now suppose the question had instead given \(EC = 10\) cm and asked whether \(DE \parallel BC\). Then we could check:

$$\frac{AD}{DB}=\frac{4}{8}=\frac{1}{2}$$ $$\frac{AE}{EC}=\frac{5}{10}=\frac{1}{2}$$

Since the ratios are equal, the converse of BPT would confirm that

$$DE \parallel BC$$

11. Worked Example 4: Using whole sides

In triangle \(XYZ\), point \(M\) lies on \(XY\) and point \(N\) lies on \(XZ\). Given:

• \(XM = 6\) cm, \(XY = 15\) cm
• \(XN = 8\) cm, \(XZ = 20\) cm

Show that \(MN \parallel YZ\).

Solution:

Here, we can compare the ratios of the smaller parts to the whole sides:

$$\frac{XM}{XY}=\frac{6}{15}=\frac{2}{5}$$ $$\frac{XN}{XZ}=\frac{8}{20}=\frac{2}{5}$$

Since

$$\frac{XM}{XY}=\frac{XN}{XZ},$$

the points divide the two sides proportionally.

Therefore, by the Converse of the Basic Proportionality Theorem,

$$MN \parallel YZ$$

If you want, you can also check using divided parts:

Since \(XY = 15\) and \(XM = 6\),

$$MY = 15 - 6 = 9$$

Since \(XZ = 20\) and \(XN = 8\),

$$NZ = 20 - 8 = 12$$

Now compare:

$$\frac{XM}{MY}=\frac{6}{9}=\frac{2}{3}$$ $$\frac{XN}{NZ}=\frac{8}{12}=\frac{2}{3}$$

The ratios are equal, so again we get

$$MN \parallel YZ$$

12. Common mistakes to avoid

• Mixing the order of ratios: If you write \(AD/DB\), then you must compare it with \(AE/EC\), not \(EC/AE\).
• Comparing wrong sides: The points must lie on two sides of the same triangle.
• Using addition instead of ratio: The theorem is about proportional division, not equal differences.
• Forgetting the converse is used to prove parallel lines: This theorem helps conclude that a line is parallel after checking the ratios.

13. Quick comparison: BPT and its converse

• BPT: Parallel line given \(\rightarrow\) ratios become equal.
• Converse of BPT: Ratios are equal \(\rightarrow\) line is parallel.

So always notice what the question gives you first.

14. When should you use this theorem?

You should think of the converse of BPT when:

• a figure shows a triangle with a segment joining two sides,
• lengths on those two sides are given,
• you are asked to prove that the joining segment is parallel to the third side.

15. Practice questions

1. In triangle \(ABC\), \(D\) lies on \(AB\) and \(E\) lies on \(AC\). If \(AD = 2\) cm, \(DB = 3\) cm, \(AE = 4\) cm, and \(EC = 6\) cm, is \(DE \parallel BC\)?
2. In triangle \(PQR\), \(S\) lies on \(PQ\) and \(T\) lies on \(PR\). If \(PS = 5\) cm, \(SQ = 10\) cm, \(PT = 4\) cm, and \(TR = 8\) cm, prove whether \(ST \parallel QR\).
3. In triangle \(XYZ\), \(M\) lies on \(XY\) and \(N\) lies on \(XZ\). If \(XM = 3\) cm, \(MY = 6\) cm, \(XN = 5\) cm, find \(NZ\) so that \(MN \parallel YZ\).

16. Answers to practice questions

1.

$$\frac{AD}{DB}=\frac{2}{3}, \quad \frac{AE}{EC}=\frac{4}{6}=\frac{2}{3}$$

The ratios are equal, so

$$DE \parallel BC$$

2.

$$\frac{PS}{SQ}=\frac{5}{10}=\frac{1}{2}, \quad \frac{PT}{TR}=\frac{4}{8}=\frac{1}{2}$$

The ratios are equal, so

$$ST \parallel QR$$

3.

For \(MN \parallel YZ\), we need:

$$\frac{XM}{MY}=\frac{XN}{NZ}$$ $$\frac{3}{6}=\frac{5}{NZ}$$ $$\frac{1}{2}=\frac{5}{NZ}$$ $$NZ = 10 \text{ cm}$$

17. Summary

The Converse of the Basic Proportionality Theorem says that if a line cuts two sides of a triangle in the same ratio, then that line is parallel to the third side.

In symbols, if in triangle \(ABC\), points \(D\) and \(E\) lie on \(AB\) and \(AC\) such that

$$\frac{AD}{DB}=\frac{AE}{EC},$$

then

$$DE \parallel BC$$

This theorem is especially useful when you are given lengths and asked to prove parallel lines. Always check that you are comparing corresponding parts in the correct order.

AA and AAA Similarity Criteria

When two figures have the same shape but may have different sizes, they are called similar. In this lesson, we will focus on similar triangles.

Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio. This means one triangle is a scaled-up or scaled-down version of the other.

The two similarity criteria in this lesson are AA and AAA. These help us prove that triangles are similar without knowing the side lengths first.

1. What does AA similarity mean?

AA stands for Angle-Angle. If two angles of one triangle are equal to two corresponding angles of another triangle, then the triangles are similar.

For example, if in triangles \(\triangle ABC\) and \(\triangle DEF\), we know:

$$ \angle A = \angle D \quad \text{and} \quad \angle B = \angle E, $$

then we can conclude:

$$ \triangle ABC \sim \triangle DEF. $$

This works because the angles in any triangle add up to \(180^\circ\). So if two pairs of angles match, the third pair must also match automatically.

For instance, if:

$$ \angle A = 50^\circ, \quad \angle B = 60^\circ, $$

then the third angle is:

$$ \angle C = 180^\circ - 50^\circ - 60^\circ = 70^\circ. $$

If another triangle has angles \(50^\circ\) and \(60^\circ\), its third angle must also be \(70^\circ\).

So AA is enough. We do not need all three angles to prove similarity.

2. What does AAA similarity mean?

AAA stands for Angle-Angle-Angle. If all three corresponding angles of two triangles are equal, then the triangles are similar.

For example, if:

$$ \angle A = \angle D, \quad \angle B = \angle E, \quad \angle C = \angle F, $$

then:

$$ \triangle ABC \sim \triangle DEF. $$

However, in practice, AAA is really just a longer version of AA. Since two equal angles already force the third angle to be equal, most proofs use AA similarity.

Important idea: AAA proves triangles are the same shape, but not necessarily the same size. One triangle could be an enlargement of the other.

3. Why do angle matches prove similarity?

If two triangles have the same angle measures, then their shape is fixed. The only thing that can change is the size. So one triangle must be a scaled copy of the other.

That is why equal corresponding angles lead to similar triangles.

4. Corresponding angles and correct order

When writing a similarity statement, the order of the letters matters. It tells us which angles and sides match.

If:

$$ \triangle ABC \sim \triangle DEF, $$

then the correspondence is:

• \(\angle A \leftrightarrow \angle D\)
• \(\angle B \leftrightarrow \angle E\)
• \(\angle C \leftrightarrow \angle F\)

And the corresponding sides are:

• \(AB \leftrightarrow DE\)
• \(BC \leftrightarrow EF\)
• \(AC \leftrightarrow DF\)

If the order is written incorrectly, then the side matching will also be wrong.

5. Angle facts often used with AA similarity

Many questions do not directly tell you that two angles are equal. You may need to find them using angle facts you already know.

• Vertically opposite angles are equal.
• Alternate interior angles are equal when lines are parallel.
• Corresponding angles are equal when lines are parallel.
• Angles in a triangle add up to \(180^\circ\).
• Angles on a straight line add up to \(180^\circ\).

These facts are very useful when proving triangles similar.

6. What can we do after proving triangles are similar?

Once triangles are similar, we can conclude:

• corresponding angles are equal,
• corresponding sides are proportional,
• missing side lengths can be found,
• geometric relationships can be proved.

If \(\triangle ABC \sim \triangle DEF\), then:

$$ \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}. $$

This is one of the main reasons similarity is so useful.

Worked Example 1: Direct use of AA

In \(\triangle PQR\), \(\angle P = 40^\circ\) and \(\angle Q = 75^\circ\). In \(\triangle XYZ\), \(\angle X = 40^\circ\) and \(\angle Y = 75^\circ\). Show that the triangles are similar.

Step 1: Compare two pairs of angles.

$$ \angle P = \angle X = 40^\circ $$ $$ \angle Q = \angle Y = 75^\circ $$

Step 2: Apply AA similarity.

Since two corresponding angles are equal, the triangles are similar:

$$ \triangle PQR \sim \triangle XYZ. $$

Step 3: Identify the correspondence.

• \(P \leftrightarrow X\)
• \(Q \leftrightarrow Y\)
• \(R \leftrightarrow Z\)

Worked Example 2: Using the third angle

Triangle \(ABC\) has angles \(52^\circ\), \(68^\circ\), and \(60^\circ\). Triangle \(DEF\) has angles \(52^\circ\), \(60^\circ\), and \(68^\circ\). Are the triangles similar?

Yes, they are similar, because all three angles match.

But we must match them in the correct order.

Let us pair equal angles:

• \(\angle A = 52^\circ \leftrightarrow \angle D = 52^\circ\)
• \(\angle B = 68^\circ \leftrightarrow \angle F = 68^\circ\)
• \(\angle C = 60^\circ \leftrightarrow \angle E = 60^\circ\)

So the correct similarity statement is:

$$ \triangle ABC \sim \triangle DFE. $$

Notice that writing \(\triangle ABC \sim \triangle DEF\) would be wrong, because the angles would not line up correctly.

Worked Example 3: Finding similarity from parallel lines

Suppose \(DE\) is parallel to \(BC\) in triangle \(ABC\), where \(D\) lies on \(AB\) and \(E\) lies on \(AC\). Show that \(\triangle ADE\) is similar to \(\triangle ABC\).

Step 1: Use angle facts from parallel lines.

Since \(DE \parallel BC\):

• \(\angle ADE = \angle ABC\) by corresponding angles
• \(\angle AED = \angle ACB\) by corresponding angles

Step 2: Apply AA similarity.

Two pairs of corresponding angles are equal, so:

$$ \triangle ADE \sim \triangle ABC. $$

Step 3: Write the side ratios.

The corresponding sides are:

• \(AD \leftrightarrow AB\)
• \(AE \leftrightarrow AC\)
• \(DE \leftrightarrow BC\)

Therefore:

$$ \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}. $$

This type of result is very common in geometry.

Worked Example 4: Use similarity to find a missing side

Triangles \(ABC\) and \(DEF\) are similar. Suppose:

$$ \angle A = \angle D, \quad \angle B = \angle E $$

and the side lengths are:

$$ AB = 6, \quad BC = 9, \quad DE = 10. $$

Find \(EF\).

Step 1: Write the similarity statement.

Since \(\angle A = \angle D\) and \(\angle B = \angle E\), we have:

$$ \triangle ABC \sim \triangle DEF. $$

So:

• \(AB \leftrightarrow DE\)
• \(BC \leftrightarrow EF\)

Step 2: Set up a proportion.

$$ \frac{AB}{DE} = \frac{BC}{EF} $$

Substitute the values:

$$ \frac{6}{10} = \frac{9}{EF} $$

Step 3: Solve.

$$ 6 \cdot EF = 10 \cdot 9 $$ $$ 6EF = 90 $$ $$ EF = 15 $$

So the missing side is:

$$ EF = 15. $$

7. Common mistakes to avoid

• Do not match angles in the wrong order. The order in the similarity statement matters.
• Do not confuse similar with congruent. Similar triangles have the same shape, but not necessarily the same size.
• Do not think AAA means same size. It only guarantees the same shape.
• Do not skip the reason. If angles are equal because of parallel lines or another fact, say why.
• Do not mix up corresponding sides. Always match sides opposite equal angles.

8. A simple method for proving triangles similar using AA

1. Find two pairs of equal angles.
2. State the reason the angles are equal.
3. Conclude that the triangles are similar by AA.
4. Write the triangles in the correct order.
5. Use corresponding sides if you need to solve for lengths.

9. Why this matters

AA similarity appears often in geometry because angles are usually easier to find than side lengths. It helps us compare triangles inside larger diagrams, especially when parallel lines or intersecting lines are involved.

This idea is also important later when studying right triangles and proofs such as the Pythagorean theorem, where smaller triangles inside a larger triangle are often shown to be similar.

Summary

AA similarity means that if two pairs of corresponding angles in two triangles are equal, then the triangles are similar.

AAA similarity also proves triangles are similar, but it is not really a separate need because once two angles match, the third must match too.

After proving triangles are similar, you can match corresponding sides and write proportions such as:

$$ \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}. $$

The key skills are finding equal angles, writing the triangles in the correct order, and using proportional sides carefully.

Lesson: SSS and SAS Similarity Criteria

In geometry, two figures are similar if they have the same shape but not necessarily the same size. For triangles, this means their corresponding angles are equal and their corresponding sides are in the same ratio.

Similarity is very useful because once we know two triangles are similar, we can compare side lengths, find missing measures, and explain why geometric relationships are true.

In this lesson, we will focus on two important ways to prove that triangles are similar:

• SSS Similarity: all three pairs of corresponding sides are proportional.
• SAS Similarity: two pairs of corresponding sides are proportional, and the included angle between them is equal.

Before we begin: what does “corresponding” mean?

Corresponding sides are sides that match in the same relative position in the two triangles. Corresponding angles are angles that match in the same relative position.

For example, if \(\triangle ABC \sim \triangle DEF\), then the order tells us the matching parts:

• \(A \leftrightarrow D\)
• \(B \leftrightarrow E\)
• \(C \leftrightarrow F\)

So the corresponding sides are:

• \(AB \leftrightarrow DE\)
• \(BC \leftrightarrow EF\)
• \(AC \leftrightarrow DF\)

If the triangles are similar, then:

$$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$

1. SSS Similarity Criterion

The letters SSS stand for Side-Side-Side.

If the three pairs of corresponding sides of two triangles are proportional, then the triangles are similar.

In symbols, if for triangles \(\triangle ABC\) and \(\triangle DEF\),

$$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$

then

$$\triangle ABC \sim \triangle DEF$$

This works because if all three side lengths grow or shrink by the same scale factor, the shape stays the same.

How to check SSS similarity

1. Match the corresponding sides correctly.
2. Write the three side ratios.
3. See whether all three ratios are equal.
4. If they are equal, the triangles are similar.

Important: The sides must be matched in the correct order. Using the wrong pairs can lead to the wrong conclusion.

2. SAS Similarity Criterion

The letters SAS stand for Side-Angle-Side.

If two pairs of corresponding sides are proportional and the included angle between those sides is equal, then the triangles are similar.

The included angle is the angle formed between the two sides you are comparing.

For example, in triangles \(\triangle ABC\) and \(\triangle DEF\), if:

• \(\frac{AB}{DE} = \frac{AC}{DF}\)
• \(\angle A = \angle D\)

and \(\angle A\) and \(\angle D\) are the angles between those pairs of sides, then:

$$\triangle ABC \sim \triangle DEF$$

How to check SAS similarity

1. Choose two sides from one triangle and the matching two sides from the other triangle.
2. Check whether the two side ratios are equal.
3. Make sure the angle between those sides is equal in both triangles.
4. If both conditions are true, the triangles are similar.

Important: The angle must be the included angle. If the equal angle is not between the two compared sides, SAS similarity does not apply.

SSS vs. SAS

• Use SSS when you know all three side lengths.
• Use SAS when you know two side lengths and the angle between them.

Worked Example 1: Using SSS Similarity

Determine whether the triangles are similar.

Triangle 1 has side lengths \(3\), \(4\), and \(5\).

Triangle 2 has side lengths \(6\), \(8\), and \(10\).

Step 1: Write the side ratios.

$$\frac{3}{6} = \frac{1}{2}, \quad \frac{4}{8} = \frac{1}{2}, \quad \frac{5}{10} = \frac{1}{2}$$

Step 2: Compare the ratios.

All three ratios are equal.

Conclusion:

$$\triangle 1 \sim \triangle 2$$

The scale factor from Triangle 1 to Triangle 2 is \(2\), because each side was multiplied by \(2\).

Worked Example 2: SSS Similarity that does not work

Determine whether triangles with side lengths \(4, 6, 9\) and \(8, 12, 16\) are similar.

Step 1: Write the ratios of corresponding sides.

$$\frac{4}{8} = \frac{1}{2}, \quad \frac{6}{12} = \frac{1}{2}, \quad \frac{9}{16} \neq \frac{1}{2}$$

Step 2: Compare the ratios.

The first two ratios are equal, but the third is not.

Conclusion:

The triangles are not similar.

This example shows why, for SSS similarity, all three pairs of sides must be proportional.

Worked Example 3: Using SAS Similarity

Determine whether the triangles are similar.

In \(\triangle ABC\), \(AB = 6\), \(AC = 9\), and \(\angle A = 50^\circ\).

In \(\triangle DEF\), \(DE = 4\), \(DF = 6\), and \(\angle D = 50^\circ\).

Step 1: Compare the two side ratios.

$$\frac{AB}{DE} = \frac{6}{4} = \frac{3}{2}$$ $$\frac{AC}{DF} = \frac{9}{6} = \frac{3}{2}$$

The two side ratios are equal.

Step 2: Compare the included angles.

\(\angle A = 50^\circ\) and \(\angle D = 50^\circ\), so the included angles are equal.

Conclusion:

$$\triangle ABC \sim \triangle DEF$$

This is true by SAS similarity.

Worked Example 4: Finding a missing side using similarity

Suppose \(\triangle ABC \sim \triangle DEF\). You know:

• \(AB = 8\)
• \(AC = 12\)
• \(DE = 10\)
• \(DF = x\)

Find \(x\).

Step 1: Match corresponding sides.

From \(\triangle ABC \sim \triangle DEF\), we have:

• \(AB \leftrightarrow DE\)
• \(AC \leftrightarrow DF\)

So:

$$\frac{AB}{DE} = \frac{AC}{DF}$$

Step 2: Substitute the known values.

$$\frac{8}{10} = \frac{12}{x}$$

Step 3: Solve by cross multiplication.

$$8x = 120$$ $$x = 15$$

Conclusion:

\(DF = 15\).

Common mistakes to avoid

• Mixing up corresponding sides: Always match sides in the same relative position.
• Using only two sides for SSS: SSS requires all three pairs of sides.
• Using the wrong angle in SAS: The equal angle must be between the two compared sides.
• Assuming equal side lengths mean congruent or similar automatically: You must check the correct condition carefully.

How to decide which criterion to use

When solving a problem, ask yourself:

• Do I know all three side lengths in both triangles? Use SSS.
• Do I know two sides and the angle between them in both triangles? Use SAS.

Why similarity matters

Once triangles are proven similar, we can do much more than just say they have the same shape. We can:

• find missing side lengths,
• compare scale factors,
• solve real-life measurement problems,
• and build important proofs in geometry.

Later in geometry, triangle similarity helps explain many results, including relationships that lead to proofs of the Pythagorean theorem.

Quick Check

1. Are triangles with side lengths \(5, 7, 9\) and \(10, 14, 18\) similar?
2. If two triangles have side ratios \(\frac{4}{6}\) and \(\frac{10}{15}\), and the included angles are both \(35^\circ\), are they similar?
3. Why is the included angle important in SAS similarity?

Answers:

1. Yes, because all side ratios are \(\frac{1}{2}\), so SSS similarity applies.
2. Yes, because \(\frac{4}{6} = \frac{10}{15} = \frac{2}{3}\) and the included angles are equal, so SAS similarity applies.
3. Because the angle must connect the two proportional sides. Without that, the triangles may not have the same shape.

Summary

Two triangles are similar if they have the same shape. The SSS similarity criterion says triangles are similar when all three pairs of corresponding sides are proportional. The SAS similarity criterion says triangles are similar when two pairs of corresponding sides are proportional and the included angle between them is equal.

To use these criteria correctly, always match corresponding sides carefully and check the right angle in SAS. Once triangles are proven similar, you can use proportional side lengths to solve many geometry problems.

Ratio of Areas of Similar Triangles

When two triangles are similar, they have the same shape but may have different sizes. Their corresponding angles are equal, and their corresponding sides are in the same ratio.

In this lesson, we will learn an important result: if two triangles are similar, then the ratio of their areas is equal to the square of the ratio of any pair of corresponding sides.

This idea is very useful because it connects shape, side lengths, and area. Once you know the scale factor between similar triangles, you can quickly compare their areas.

1. Review: What are similar triangles?

Two triangles are similar if:

• their corresponding angles are equal, and
• their corresponding sides are proportional.

For example, if \(\triangle ABC \sim \triangle DEF\), then the vertices match in order:

• \(A \leftrightarrow D\)
• \(B \leftrightarrow E\)
• \(C \leftrightarrow F\)

So the corresponding sides are:

• \(AB \leftrightarrow DE\)
• \(BC \leftrightarrow EF\)
• \(AC \leftrightarrow DF\)

This means:

$$ \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} $$

This common ratio is called the scale factor.

2. The main theorem

If two triangles are similar, then the ratio of their areas is the square of the ratio of their corresponding sides.

If \(\triangle ABC \sim \triangle DEF\), then:

$$ \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{AC}{DF}\right)^2 $$

So if the side ratio is \(m:n\), then the area ratio is:

$$ m^2:n^2 $$

3. Why do we square the side ratio?

Area depends on two measurements. For a triangle,

$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$

In similar triangles, not only the sides but also the heights are in the same ratio.

Suppose the scale factor from one triangle to another is \(k\). Then:

• the base becomes \(k\) times as large, and
• the height also becomes \(k\) times as large.

So the area becomes:

$$ \frac{1}{2} \times (k \cdot \text{base}) \times (k \cdot \text{height}) = k^2 \left(\frac{1}{2} \times \text{base} \times \text{height}\right) $$

That is why the area ratio is \(k^2\), not just \(k\).

4. Simple proof of the theorem

Let two similar triangles be \(\triangle ABC\) and \(\triangle DEF\), with corresponding sides in the ratio

$$ \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = k $$

Take \(AB\) and \(DE\) as corresponding bases. Let their corresponding heights be \(h_1\) and \(h_2\).

Because the triangles are similar, the heights are also in the same ratio:

$$ \frac{h_1}{h_2} = k $$

Now use the area formula:

$$ \text{Area of } \triangle ABC = \frac{1}{2} \cdot AB \cdot h_1 $$ $$ \text{Area of } \triangle DEF = \frac{1}{2} \cdot DE \cdot h_2 $$

Divide the two areas:

$$ \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{\frac{1}{2} \cdot AB \cdot h_1}{\frac{1}{2} \cdot DE \cdot h_2} = \frac{AB}{DE} \cdot \frac{h_1}{h_2} $$

Since both ratios equal \(k\),

$$ \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = k \cdot k = k^2 $$

Therefore,

$$ \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \left(\frac{AB}{DE}\right)^2 $$

5. Important rule to remember

• If the ratio of corresponding sides is \(a:b\), then the ratio of areas is \(a^2:b^2\).
• If the ratio of areas is \(a:b\), then the ratio of corresponding sides is \(\sqrt{a}:\sqrt{b}\).

This second rule is also very important. To go from side ratio to area ratio, square. To go from area ratio to side ratio, take the square root.

6. Worked Example 1: Finding the area ratio from the side ratio

Two similar triangles have corresponding sides in the ratio \(2:3\). Find the ratio of their areas.

Step 1: Write the side ratio.

$$ 2:3 $$

Step 2: Square both parts.

$$ 2^2:3^2 = 4:9 $$

Answer: The ratio of their areas is \(4:9\).

Worked Example 2: Using actual side lengths

\(\triangle PQR \sim \triangle XYZ\). If \(PQ = 6\text{ cm}\) and \(XY = 9\text{ cm}\), find the ratio of the areas of the two triangles.

Since the triangles are similar, corresponding sides are proportional. The side ratio is:

$$ \frac{PQ}{XY} = \frac{6}{9} = \frac{2}{3} $$

Now square the ratio:

$$ \frac{\text{Area of } \triangle PQR}{\text{Area of } \triangle XYZ} = \left(\frac{2}{3}\right)^2 = \frac{4}{9} $$

Answer:

$$ \text{Area of } \triangle PQR : \text{Area of } \triangle XYZ = 4:9 $$

Worked Example 3: Finding an unknown area

Two similar triangles have corresponding sides in the ratio \(3:5\). If the area of the smaller triangle is \(27\text{ cm}^2\), find the area of the larger triangle.

Step 1: Find the area ratio.

$$ 3:5 \Rightarrow 3^2:5^2 = 9:25 $$

So,

$$ \text{smaller area} : \text{larger area} = 9:25 $$

Step 2: Use the known area.

If \(9\) parts correspond to \(27\text{ cm}^2\), then 1 part is:

$$ 27 \div 9 = 3\text{ cm}^2 $$

Step 3: Find 25 parts.

$$ 25 \times 3 = 75\text{ cm}^2 $$

Answer: The area of the larger triangle is \(75\text{ cm}^2\).

Worked Example 4: Finding the side ratio from the area ratio

The areas of two similar triangles are in the ratio \(16:81\). Find the ratio of their corresponding sides.

To go from area ratio to side ratio, take the square root of each term:

$$ \sqrt{16}:\sqrt{81} = 4:9 $$

Answer: The ratio of corresponding sides is \(4:9\).

7. A geometric interpretation

Imagine enlarging a triangle by a scale factor of \(2\). Every side doubles. Every height also doubles. Since area uses base and height, the new area becomes:

$$ 2^2 = 4 $$

times the original area.

If a triangle is enlarged by a scale factor of \(3\), then its area becomes:

$$ 3^2 = 9 $$

times the original area.

If a triangle is reduced by a scale factor of \(\frac{1}{2}\), then its area becomes:

$$ \left(\frac{1}{2}\right)^2 = \frac{1}{4} $$

of the original area.

This helps you see that area changes much faster than side length.

8. Common mistakes to avoid

• Do not use the side ratio directly as the area ratio. If side ratio is \(2:3\), area ratio is not \(2:3\). It is \(4:9\).
• Make sure the triangles are similar. This rule works only for similar triangles.
• Match corresponding sides correctly. If you compare the wrong sides, your ratio will be wrong.
• When given areas, take square roots to find side ratios. Do not square again.

9. Quick practice ideas

1. If the side ratio is \(4:7\), what is the area ratio?
2. If the area ratio is \(25:36\), what is the side ratio?
3. Two similar triangles have side ratio \(5:8\). If the smaller area is \(50\text{ cm}^2\), find the larger area.

Answers:

1. \(16:49\)
2. \(5:6\)
3. Area ratio \(= 25:64\). If 25 parts = 50, then 1 part = 2, so 64 parts = \(128\text{ cm}^2\).

10. Summary

For similar triangles, corresponding sides are in the same ratio, and corresponding heights are also in that ratio. Since area is based on both base and height, the ratio of the areas is the square of the ratio of corresponding sides.

So remember:

$$ \frac{\text{Area of one triangle}}{\text{Area of the other triangle}} = \left(\frac{\text{corresponding side}}{\text{corresponding side}}\right)^2 $$

This means:

• side ratio \(a:b\) gives area ratio \(a^2:b^2\),
• and area ratio \(a:b\) gives side ratio \(\sqrt{a}:\sqrt{b}\).

Once you understand this rule, you can solve many geometry problems involving similar figures more quickly and accurately.

Similarity in Right-Angled Triangles

In this lesson, you will learn what happens when an altitude is drawn from the right angle of a right-angled triangle to the hypotenuse.

This creates two smaller right-angled triangles inside the original triangle. A very important fact is that all three triangles are similar. That means they have the same shape, even though they may have different sizes.

Once we know the triangles are similar, we can compare matching sides and find many useful relationships. These relationships help us solve geometry problems and even connect to the Pythagorean theorem.

1. The main figure

Consider right-angled triangle \(\triangle ABC\), where the right angle is at \(C\). This means:

$$\angle C = 90^\circ$$

The side opposite the right angle, \(AB\), is the hypotenuse.

Now draw altitude \(CD\) from point \(C\) straight down to the hypotenuse \(AB\). Point \(D\) lies on \(AB\), and:

$$CD \perp AB$$

This divides the original triangle into two smaller triangles:

• \(\triangle ACD\)
• \(\triangle CBD\)

So now we have three right-angled triangles in the figure:

• the large triangle \(\triangle ABC\)
• the small triangle \(\triangle ACD\)
• the small triangle \(\triangle CBD\)

2. Why the triangles are similar

To show triangles are similar, we usually compare angles.

First, notice that \(\triangle ABC\) is right-angled at \(C\), and both smaller triangles are right-angled at \(D\), because the altitude is perpendicular to the hypotenuse.

So:

$$\angle C = \angle CDA = \angle CDB = 90^\circ$$

Now look at shared angles:

• \(\triangle ACD\) shares angle \(A\) with \(\triangle ABC\)
• \(\triangle CBD\) shares angle \(B\) with \(\triangle ABC\)

That gives us enough information to use AA similarity (Angle-Angle similarity).

Therefore:

$$\triangle ABC \sim \triangle ACD \sim \triangle CBD$$

3. Matching sides carefully

When triangles are similar, the most important step is to match the correct sides. Let us label the sides of the large triangle:

• \(AB = c\) is the hypotenuse
• \(AC = b\)
• \(BC = a\)

The altitude \(CD\) splits the hypotenuse into two parts:

• \(AD = p\)
• \(DB = q\)

So the whole hypotenuse is:

$$AB = AD + DB$$

$$c = p + q$$

4. Important similarity relationships

Because the triangles are similar, several useful formulas appear.

(a) Leg-hypotenuse projection relationships

Each leg of the large triangle is the geometric mean of the hypotenuse and the part of the hypotenuse next to that leg.

For leg \(AC\):

$$AC^2 = AB \cdot AD$$

$$b^2 = c p$$

For leg \(BC\):

$$BC^2 = AB \cdot DB$$

$$a^2 = c q$$

(b) Altitude relationship

The altitude is the geometric mean of the two segments of the hypotenuse.

$$CD^2 = AD \cdot DB$$

$$h^2 = p q$$

Here \(h = CD\).

(c) The hypotenuse is split into two parts

$$c = p + q$$

These three ideas are the key results for this topic.

5. Where these formulas come from

Let us see how one of them is formed using similar triangles.

Compare \(\triangle ABC\) and \(\triangle ACD\).

The matching sides are:

• hypotenuse \(AB\) of the large triangle matches hypotenuse \(AC\) of the smaller triangle
• side \(AC\) of the large triangle matches side \(AD\) of the smaller triangle

So we can write:

$$\frac{AB}{AC} = \frac{AC}{AD}$$

Cross-multiply:

$$AC^2 = AB \cdot AD$$

This is:

$$b^2 = c p$$

In the same way, comparing \(\triangle ABC\) and \(\triangle CBD\) gives:

$$BC^2 = AB \cdot DB$$

And comparing the two smaller triangles gives:

$$CD^2 = AD \cdot DB$$

6. Connection to the Pythagorean theorem

These similarity results lead directly to the Pythagorean theorem.

We already know:

$$AC^2 = AB \cdot AD$$

$$BC^2 = AB \cdot DB$$

Add the equations:

$$AC^2 + BC^2 = AB \cdot AD + AB \cdot DB$$

Factor out \(AB\):

$$AC^2 + BC^2 = AB(AD + DB)$$

But \(AD + DB = AB\), so:

$$AC^2 + BC^2 = AB \cdot AB$$

$$AC^2 + BC^2 = AB^2$$

This is the Pythagorean theorem for right-angled triangle \(\triangle ABC\).

7. Worked Example 1: Finding a leg from the hypotenuse segment

In a right-angled triangle, the hypotenuse is \(25\) cm. The altitude from the right angle divides the hypotenuse so that \(AD = 9\) cm and \(DB = 16\) cm.

Find the two legs.

Step 1: Use the leg formulas

$$AC^2 = AB \cdot AD$$

$$AC^2 = 25 \cdot 9 = 225$$

$$AC = 15 \text{ cm}$$

Now for the other leg:

$$BC^2 = AB \cdot DB$$

$$BC^2 = 25 \cdot 16 = 400$$

$$BC = 20 \text{ cm}$$

Answer: The legs are \(15\) cm and \(20\) cm.

Check:

$$15^2 + 20^2 = 225 + 400 = 625 = 25^2$$

The answer fits the Pythagorean theorem.

8. Worked Example 2: Finding the altitude

Using the same triangle, find the altitude \(CD\).

We use:

$$CD^2 = AD \cdot DB$$

$$CD^2 = 9 \cdot 16 = 144$$

$$CD = 12 \text{ cm}$$

Answer: The altitude is \(12\) cm.

9. Worked Example 3: Finding a missing hypotenuse segment

In right-angled triangle \(ABC\), \(AC = 8\) cm and \(AB = 10\) cm. The altitude from \(C\) meets hypotenuse \(AB\) at \(D\).

Find \(AD\).

Use:

$$AC^2 = AB \cdot AD$$

Substitute the values:

$$8^2 = 10 \cdot AD$$

$$64 = 10AD$$

$$AD = 6.4 \text{ cm}$$

Answer: \(AD = 6.4\) cm.

If needed, we can find \(DB\) too:

$$DB = AB - AD = 10 - 6.4 = 3.6 \text{ cm}$$

10. Worked Example 4: Using similarity to solve a fuller problem

In a right-angled triangle, the altitude to the hypotenuse divides the hypotenuse into segments of lengths \(4\) cm and \(5\) cm.

Find:

1. the hypotenuse,
2. the altitude,
3. the two legs.

Step 1: Find the hypotenuse

$$AB = AD + DB = 4 + 5 = 9 \text{ cm}$$

Step 2: Find the altitude

$$CD^2 = AD \cdot DB$$

$$CD^2 = 4 \cdot 5 = 20$$

$$CD = \sqrt{20} = 2\sqrt{5} \text{ cm}$$

Step 3: Find the legs

For the leg next to \(AD\):

$$AC^2 = AB \cdot AD = 9 \cdot 4 = 36$$

$$AC = 6 \text{ cm}$$

For the leg next to \(DB\):

$$BC^2 = AB \cdot DB = 9 \cdot 5 = 45$$

$$BC = \sqrt{45} = 3\sqrt{5} \text{ cm}$$

Answer:

• Hypotenuse: \(9\) cm
• Altitude: \(2\sqrt{5}\) cm
• Legs: \(6\) cm and \(3\sqrt{5}\) cm

11. Common mistakes to avoid

• Mixing up the segments of the hypotenuse. Remember that \(AD\) and \(DB\) are only parts of the hypotenuse, not the whole hypotenuse.
• Using the wrong formula for a side. For example, \(AC^2 = AB \cdot AD\), not \(AB \cdot DB\).
• Forgetting to square root at the end. If you find \(AC^2 = 49\), then \(AC = 7\), not \(49\).
• Matching the wrong sides in similar triangles. Always compare angles first so you know which sides correspond.

12. Quick guide to remember

In a right-angled triangle with altitude to the hypotenuse:

• The three triangles are similar.
• The hypotenuse is split into two parts: \(c = p + q\).
• Each leg squared equals the hypotenuse times its nearby segment:

$$b^2 = cp$$

$$a^2 = cq$$

• The altitude squared equals the product of the two hypotenuse segments:

$$h^2 = pq$$

13. Brief summary

When an altitude is drawn from the right angle to the hypotenuse of a right-angled triangle, it creates two smaller triangles that are similar to the original triangle and to each other.

This similarity gives important relationships between the legs, the altitude, and the two parts of the hypotenuse. These relationships are powerful tools for solving problems and help explain why the Pythagorean theorem is true.

Proof of the Pythagorean Theorem via Similarity

The Pythagorean Theorem is one of the most important results in geometry. It says that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

If a right triangle has legs of lengths \(a\) and \(b\), and hypotenuse of length \(c\), then the theorem states:

$$a^2 + b^2 = c^2$$

In this lesson, we will prove this theorem using similar triangles. This is a powerful idea because it shows that the theorem is not just a rule to memorize. It comes from geometric relationships inside the triangle itself.

Goal of the lesson: Understand how drawing an altitude in a right triangle creates smaller triangles that are similar to the original triangle, and then use those similarity relationships to prove \(a^2 + b^2 = c^2\).

1. Review: What does it mean for triangles to be similar?

Two triangles are similar if they have the same shape, even if they are different sizes. In similar triangles:

• Corresponding angles are equal.
• Corresponding side lengths are proportional.

For example, if two triangles are similar, then ratios of matching sides are equal, such as

$$\frac{\text{side 1 of triangle A}}{\text{matching side 1 of triangle B}} = \frac{\text{side 2 of triangle A}}{\text{matching side 2 of triangle B}}$$

We will use this idea to connect side lengths in a right triangle.

2. Set up the right triangle

Consider right triangle \(\triangle ABC\), where:

• \(\angle C = 90^\circ\)
• \(AC = b\)
• \(BC = a\)
• \(AB = c\) is the hypotenuse

Now draw an altitude from point \(C\) straight down to the hypotenuse \(AB\). Let it meet \(AB\) at point \(D\).

This creates two smaller right triangles inside the original triangle:

• \(\triangle ACD\)
• \(\triangle CBD\)

Let the two pieces of the hypotenuse be:

• \(AD = x\)
• \(DB = y\)

Since \(AB = c\), we also know:

$$x + y = c$$

3. Why are the triangles similar?

We now compare the three triangles:

• the large triangle \(\triangle ABC\)
• the small triangle \(\triangle ACD\)
• the small triangle \(\triangle CBD\)

Each small triangle has a right angle at \(D\), and each shares one acute angle with the original triangle. Because they have two matching angles, we can use the AA similarity rule.

So we get:

$$\triangle ABC \sim \triangle ACD \sim \triangle CBD$$

This is the key idea of the proof. Once we know the triangles are similar, we can write proportions between matching sides.

4. First similarity relationship

Compare \(\triangle ABC\) and \(\triangle ACD\).

In these similar triangles:

• The hypotenuse of the large triangle is \(AB = c\).
• The side \(AC = b\) in the large triangle matches \(AD = x\) in the smaller triangle.
• The side \(AC = b\) in the small triangle matches \(AB = c\) in the large triangle.

This gives the proportion:

$$\frac{b}{c} = \frac{x}{b}$$

Now multiply across:

$$b^2 = cx$$

This is our first useful equation.

5. Second similarity relationship

Now compare \(\triangle ABC\) and \(\triangle CBD\).

In these similar triangles:

• The hypotenuse of the large triangle is \(AB = c\).
• The leg \(BC = a\) of the large triangle matches \(DB = y\) in the smaller triangle.
• The leg \(BC = a\) in the small triangle matches \(AB = c\) in the large triangle.

This gives the proportion:

$$\frac{a}{c} = \frac{y}{a}$$

Multiply across:

$$a^2 = cy$$

This is our second useful equation.

6. Combine the two equations

We found:

$$a^2 = cy$$ $$b^2 = cx$$

Add the equations:

$$a^2 + b^2 = cy + cx$$

Factor out \(c\):

$$a^2 + b^2 = c(x+y)$$

But earlier we saw that \(x+y=c\). Substitute that in:

$$a^2 + b^2 = c(c)$$ $$a^2 + b^2 = c^2$$

This proves the Pythagorean Theorem using similarity.

7. Why this proof is important

This proof shows that the Pythagorean Theorem comes from the structure of a right triangle. When the altitude to the hypotenuse is drawn, the original triangle breaks into two smaller triangles that have the same shape as the original.

Because of that shared shape, the side lengths follow proportional relationships. Those proportions lead directly to the equation \(a^2 + b^2 = c^2\).

So the theorem is not just about arithmetic. It is a geometric fact based on similarity.

8. A useful side result

From the proof, we discovered:

$$a^2 = cy \quad \text{and} \quad b^2 = cx$$

These are called geometric mean relationships in a right triangle. They tell us that each leg squared equals the hypotenuse times the nearby segment of the hypotenuse.

This can help solve many geometry problems even when you are not directly proving the theorem.

9. Worked Example 1: Using the proof relationships with numbers

In a right triangle, suppose the hypotenuse is split by the altitude into segments of lengths \(4\) and \(5\). So:

$$x = 4, \quad y = 5, \quad c = 9$$

Find the lengths of the legs \(a\) and \(b\).

Step 1: Use the relationships from similarity.

$$a^2 = cy = 9 \cdot 5 = 45$$ $$b^2 = cx = 9 \cdot 4 = 36$$

Step 2: Take square roots.

$$a = \sqrt{45} = 3\sqrt{5}$$ $$b = \sqrt{36} = 6$$

Answer: The legs are \(3\sqrt{5}\) and \(6\).

Check using the Pythagorean Theorem:

$$a^2 + b^2 = 45 + 36 = 81 = 9^2$$

The result is correct.

10. Worked Example 2: Proving the theorem step by step

Suppose a student says, “I know the triangles are similar, but I do not see how that gives \(a^2+b^2=c^2\).” Let us write the proof clearly in order.

1. Start with right triangle \(\triangle ABC\), with legs \(a\) and \(b\), hypotenuse \(c\).
2. Draw altitude \(CD\) to hypotenuse \(AB\), splitting it into \(AD=x\) and \(DB=y\).
3. By AA, \(\triangle ABC \sim \triangle ACD\).
4. From similarity, $$\frac{b}{c} = \frac{x}{b}$$ so $$b^2 = cx$$
5. By AA, \(\triangle ABC \sim \triangle CBD\).
6. From similarity, $$\frac{a}{c} = \frac{y}{a}$$ so $$a^2 = cy$$
7. Add the equations: $$a^2 + b^2 = cy + cx$$
8. Factor: $$a^2 + b^2 = c(x+y)$$
9. Since \(x+y=c\), substitute: $$a^2+b^2=c\cdot c=c^2$$

Conclusion: $$a^2+b^2=c^2$$

11. Worked Example 3: Find a missing leg segment

In a right triangle, one leg has length \(8\), and the hypotenuse has length \(10\). The altitude to the hypotenuse divides the hypotenuse into segments \(x\) and \(y\), where the segment next to the leg of length \(8\) is \(x\). Find \(x\).

Step 1: Use the relationship for that leg.

If the leg is \(b=8\), then

$$b^2 = cx$$

Substitute:

$$8^2 = 10x$$ $$64 = 10x$$ $$x = 6.4$$

Answer: The segment is \(6.4\).

Check: Since the whole hypotenuse is \(10\), the other segment is

$$y = 10 - 6.4 = 3.6$$

Then the other leg should satisfy

$$a^2 = cy = 10(3.6) = 36$$ $$a = 6$$

Now check the Pythagorean Theorem:

$$6^2 + 8^2 = 36 + 64 = 100 = 10^2$$

Everything is consistent.

12. Worked Example 4: Identify and fix a common mistake

A student writes:

$$\frac{a}{c} = \frac{a}{y}$$

and concludes that \(ay = ac\), so \(y=c\).

What went wrong?

The problem is that the student matched the wrong sides. In similar triangles, you must compare corresponding sides. The side \(a\) in the large triangle does not match the side \(a\) in the smaller triangle just because they have the same letter position in your notes.

You must first identify which angles match, then match the sides opposite those angles.

The correct proportion is:

$$\frac{a}{c} = \frac{y}{a}$$

Then multiply across:

$$a^2 = cy$$

Lesson from this mistake: In similarity problems, always match sides by their positions in the triangles, not just by what seems convenient.

13. Tips for writing the proof on your own

• Start by naming the right triangle and the altitude point clearly.
• State which triangles are similar and why: usually AA similarity.
• Write one correct proportion from the first pair of triangles.
• Write a second correct proportion from the second pair.
• Simplify each into equations like \(a^2=cy\) and \(b^2=cx\).
• Add the equations and use \(x+y=c\).
• End with the final statement: $$a^2+b^2=c^2$$

14. Common misunderstandings

• Thinking all side ratios can be mixed randomly: Only corresponding sides can be compared.
• Forgetting why the triangles are similar: You must mention angle matching, usually AA.
• Forgetting that \(x+y=c\): The two smaller hypotenuse pieces add to the whole hypotenuse.
• Mixing up the legs and hypotenuse: The hypotenuse is always opposite the right angle.

15. Brief summary

To prove the Pythagorean Theorem using similarity, start with a right triangle and draw the altitude from the right angle to the hypotenuse. This creates two smaller triangles that are similar to the original triangle.

Using proportional sides, we get:

$$a^2 = cy \quad \text{and} \quad b^2 = cx$$

Adding these gives:

$$a^2+b^2 = c(x+y)$$

Since \(x+y=c\), it follows that:

$$a^2+b^2=c^2$$

That is the Pythagorean Theorem, proved using similarity.

Converse of the Pythagorean Theorem

In earlier work, you may have learned the Pythagorean Theorem: in a right triangle, if the legs have lengths \(a\) and \(b\), and the hypotenuse has length \(c\), then

$$a^2+b^2=c^2$$

The converse of a statement switches the “if” and the “then.” So the Converse of the Pythagorean Theorem says:

If the square of the longest side of a triangle equals the sum of the squares of the other two sides, then the triangle is a right triangle.

In symbols, if a triangle has side lengths \(a\), \(b\), and \(c\), where \(c\) is the longest side, and

$$a^2+b^2=c^2,$$

then the angle opposite side \(c\) is a right angle, so the triangle has a \(90^\circ\) angle.

This idea is very useful because it lets us work backward. Instead of starting with a right triangle and finding a missing side, we start with the side lengths and decide whether the triangle must be right.

This connects to triangle similarity and proofs of the Pythagorean Theorem because it gives a way to recognize when a triangle has the exact relationship that right triangles have.

Why the longest side matters

When using the converse, you must compare the longest side to the other two sides. In a right triangle, the hypotenuse is always the longest side, so in the converse, the side being tested as \(c\) must be the longest one.

For example, if the side lengths are \(5\), \(12\), and \(13\), then \(13\) is the longest side. We test whether

$$5^2+12^2=13^2$$

If it is true, the triangle is right.

How to use the converse

1. Identify the longest side. Call it \(c\).

2. Square the two shorter sides and add them: \(a^2+b^2\).

3. Square the longest side: \(c^2\).

4. Compare the results.

• If \(a^2+b^2=c^2\), the triangle is a right triangle.

• If they are not equal, then the triangle is not a right triangle.

Important note: the converse only tells us that the triangle is right when the equation is exactly true. If the sums are different, the triangle is not right.

Common Pythagorean triples

A Pythagorean triple is a set of three whole numbers that satisfy \(a^2+b^2=c^2\). These often appear in problems.

• \(3,4,5\)

• \(5,12,13\)

• \(8,15,17\)

• \(7,24,25\)

Multiples of these also work:

• \(6,8,10\) is a multiple of \(3,4,5\)

• \(9,12,15\) is also a multiple of \(3,4,5\)

If you recognize one of these sets, you can quickly tell that the triangle is right.

Worked Example 1: A basic whole-number example

Determine whether a triangle with side lengths \(6\), \(8\), and \(10\) is a right triangle.

Step 1: Identify the longest side. The longest side is \(10\).

Step 2: Test the converse.

$$6^2+8^2=36+64=100$$

$$10^2=100$$

Since

$$6^2+8^2=10^2,$$

the triangle is a right triangle.

The angle opposite the side of length \(10\) is \(90^\circ\).

Worked Example 2: A triangle that is not right

Determine whether a triangle with side lengths \(6\), \(7\), and \(9\) is a right triangle.

Step 1: The longest side is \(9\).

Step 2: Compare the squares.

$$6^2+7^2=36+49=85$$

$$9^2=81$$

Since

$$85\neq81,$$

the triangle is not a right triangle.

Even though the numbers are close, the converse requires an exact equality.

Worked Example 3: Using variables

A triangle has side lengths \(x\), \(24\), and \(25\). If the triangle is a right triangle, find \(x\).

Since \(25\) is the longest side, it must be the hypotenuse if the triangle is right. Use the converse condition:

$$x^2+24^2=25^2$$

$$x^2+576=625$$

$$x^2=49$$

$$x=7$$

So the missing side length is \(7\).

This gives the side lengths \(7\), \(24\), and \(25\), which is a Pythagorean triple.

Worked Example 4: Applying the theorem in a geometric figure

Triangle \(ABC\) has side lengths \(AB=15\), \(BC=20\), and \(AC=25\). Prove that \(\angle B\) is a right angle.

To show that \(\angle B\) is a right angle, we look at the side opposite \(\angle B\). The side opposite \(\angle B\) is \(AC\), and its length is \(25\). This is also the longest side.

Now test the converse:

$$AB^2+BC^2=15^2+20^2=225+400=625$$

$$AC^2=25^2=625$$

Since

$$AB^2+BC^2=AC^2,$$

by the Converse of the Pythagorean Theorem, triangle \(ABC\) is a right triangle.

Therefore, the angle opposite \(AC\), which is \(\angle B\), measures \(90^\circ\).

How this connects to proofs

In geometry, you often need to prove that an angle is a right angle. One powerful method is to show that the side lengths satisfy the converse of the Pythagorean Theorem.

So if a problem gives you three side lengths, or enough information to find them, you can:

• identify the longest side,

• check whether the squares of the two shorter sides add to the square of the longest side, and

• conclude that the included angle opposite the longest side is a right angle.

This is often used in coordinate geometry too. If you calculate the lengths of three sides and they satisfy the converse, then the triangle is right.

Common mistakes to avoid

• Using the wrong side as \(c\): always choose the longest side.

• Forgetting to square correctly: for example, \(12^2=144\), not \(24\).

• Assuming “close enough” is good enough: the equation must be exactly true.

• Mixing up the theorem and its converse: the original theorem starts with a right triangle; the converse starts with side lengths and proves the triangle is right.

Quick check questions

1. Are side lengths \(9\), \(12\), and \(15\) the sides of a right triangle?

2. Are side lengths \(10\), \(11\), and \(15\) the sides of a right triangle?

3. A triangle has side lengths \(8\), \(15\), and \(17\). Which angle is the right angle?

Answers

1. Yes, because \(9^2+12^2=81+144=225=15^2\).

2. No, because \(10^2+11^2=100+121=221\), and \(15^2=225\), so they are not equal.

3. The right angle is opposite the side of length \(17\).

Summary

The Converse of the Pythagorean Theorem helps you decide whether a triangle is a right triangle by using its side lengths.

If the longest side is \(c\), and the other sides are \(a\) and \(b\), then:

$$a^2+b^2=c^2 \Rightarrow \text{the triangle is a right triangle}$$

Always test the longest side, and remember that the equality must be exact. This theorem is especially useful in geometry proofs when you need to show that an angle measures \(90^\circ\).

Applications of the Pythagorean Theorem in Complex Figures

The Pythagorean Theorem is one of the most important tools in geometry. You may already know how to use it in a simple right triangle. In this lesson, you will learn how to apply it in more complex figures, such as rhombuses, trapezoids, and even 3D-looking architectural shapes.

The big idea is this: even when a figure does not look like a right triangle at first, you can often find or create a right triangle inside it. Once you do that, you can use the Pythagorean Theorem to find missing lengths.

The theorem says that in a right triangle with legs of lengths \(a\) and \(b\), and hypotenuse of length \(c\),

$$a^2+b^2=c^2$$

Remember that the hypotenuse is always the side opposite the right angle, and it is the longest side of the triangle.

Why this matters in complex figures

In many geometry problems, shapes are made from several smaller parts. A diagonal, a height, or a perpendicular segment can split a larger figure into right triangles. If you identify those triangles correctly, the Pythagorean Theorem helps you connect side lengths that are not obvious at first.

This is especially useful when working with:

• Rhombuses, where diagonals often form right triangles
• Trapezoids, where dropping heights creates right triangles
• Composite or architectural figures, where slanted edges and vertical heights form right triangles

Step-by-step strategy

1. Look for a right angle or a segment you can draw that creates one.
2. Identify the sides of the right triangle.
3. Decide which side is the hypotenuse.
4. Substitute the known values into \(a^2+b^2=c^2\).
5. Solve carefully and check whether the answer makes sense in the figure.

Important geometric facts to remember

• In a rhombus, all sides are equal.
• The diagonals of a rhombus bisect each other, meaning they cut each other into equal halves.
• In many rhombus problems, the diagonals also form right triangles inside the shape.
• In a trapezoid, drawing heights from the shorter base to the longer base often creates right triangles on the sides.
• In 3D drawings of buildings or structures, focus on the flat right triangle relationships shown in the diagram, such as height, base, and slanted roof edge.

Example 1: Rhombus with diagonals

A rhombus has diagonals of lengths 10 cm and 24 cm. Find the length of each side.

Step 1: Use properties of a rhombus.

The diagonals bisect each other, so each diagonal is cut in half.

That means the half-lengths are:

• \(10 \div 2 = 5\)
• \(24 \div 2 = 12\)

Step 2: Find the right triangle.

The half-diagonals and one side of the rhombus form a right triangle. The legs are 5 cm and 12 cm, and the side of the rhombus is the hypotenuse.

Let the side length be \(s\).

$$5^2+12^2=s^2$$ $$25+144=s^2$$ $$169=s^2$$ $$s=13$$

Answer: Each side of the rhombus is 13 cm.

What to notice: The rhombus itself is not a triangle, but its diagonals created a right triangle inside it.

Example 2: Isosceles trapezoid

An isosceles trapezoid has bases 18 m and 10 m, and each leg has length 5 m. Find the height of the trapezoid.

Step 1: Compare the bases.

The difference between the bases is

$$18-10=8$$

Because the trapezoid is isosceles, this extra length is split equally on both sides. So each small horizontal piece is

$$8 \div 2 = 4$$

Step 2: Create a right triangle.

Drop a perpendicular from one top corner to the longer base. This forms a right triangle with:

• hypotenuse = 5 m
• one leg = 4 m
• other leg = height \(h\)

Now use the Pythagorean Theorem:

$$4^2+h^2=5^2$$ $$16+h^2=25$$ $$h^2=9$$ $$h=3$$

Answer: The height of the trapezoid is 3 m.

What to notice: The trapezoid was turned into a rectangle-like middle section and two right triangles on the sides. This is a common method.

Example 3: Finding a diagonal in a rectangle inside a larger figure

A garden design is made of a rectangle attached to other shapes. The rectangle part has length 12 ft and width 9 ft. Find the diagonal across the rectangle.

Even if the rectangle is part of a bigger design, the diagonal of the rectangle forms a right triangle with the length and width.

Let the diagonal be \(d\).

$$12^2+9^2=d^2$$ $$144+81=d^2$$ $$225=d^2$$ $$d=15$$

Answer: The diagonal is 15 ft.

What to notice: In composite figures, focus on the part of the shape that gives you a right triangle. You do not always need the whole figure at once.

Example 4: 3D architectural profile

A side view of a roof forms a right triangle. The vertical height from the top of the wall to the peak is 6 m, and the horizontal distance from the wall to the peak is 8 m. Find the slanted roof length.

This is a right triangle even though it represents part of a 3D building. The slanted roof is the hypotenuse.

Let the roof length be \(r\).

$$6^2+8^2=r^2$$ $$36+64=r^2$$ $$100=r^2$$ $$r=10$$

Answer: The slanted roof length is 10 m.

What to notice: In architecture or 3D drawings, you often use the right triangle from a side view or cross-section.

When the missing side is a leg instead of the hypotenuse

Sometimes the hypotenuse is known, and you need one of the legs. In that case, rearrange the formula:

$$a^2=c^2-b^2$$

Then take the square root at the end.

For example, if a slanted side is 13 and one leg is 5, then the other leg is:

$$a^2=13^2-5^2$$ $$a^2=169-25=144$$ $$a=12$$

This is useful in trapezoids, rhombuses, and design problems where you know the longest side.

Common mistakes to avoid

• Using the theorem on a triangle that is not right. Always check for a right angle first.
• Choosing the wrong hypotenuse. The hypotenuse is opposite the right angle.
• Forgetting to halve diagonals in a rhombus. If diagonals bisect each other, use the half-lengths in the triangle.
• Using the full difference of trapezoid bases on one side. In an isosceles trapezoid, split the difference equally.
• Forgetting the square root. If you solve for \(c^2=169\), then \(c=13\), not 169.

How this connects to similarity and proofs

In geometry, the Pythagorean Theorem is often connected to similar triangles. In some proofs, a large right triangle is split into smaller triangles that are similar. Those relationships help explain why

$$a^2+b^2=c^2$$

is true.

When solving problems in complex figures, you are using that same idea in a practical way: breaking a large figure into smaller parts and using the relationships between them.

Quick practice questions

1. A rhombus has diagonals 16 cm and 30 cm. What is the side length?
2. A trapezoid has bases 20 and 12, and each leg is 5. What is its height?
3. A rectangular window is 8 ft tall and 15 ft wide. What is the length of the diagonal brace across it?

Answers

1. Half-diagonals are 8 and 15, so side \(=\sqrt{8^2+15^2}=\sqrt{289}=17\) cm.
2. Difference of bases is 8, so each side part is 4. Height \(=\sqrt{5^2-4^2}=\sqrt{9}=3\).
3. Diagonal \(=\sqrt{8^2+15^2}=\sqrt{289}=17\) ft.

Summary

The Pythagorean Theorem is not only for simple right triangles. In more complex figures, you can often find hidden right triangles by using diagonals, heights, or side views. Once you identify the right triangle correctly, you can use

$$a^2+b^2=c^2$$

to find missing lengths in rhombuses, trapezoids, rectangles, and architectural profiles. The key is to break the figure into simpler parts and work carefully.