Chapter 15

Triangle Properties, Centers, and Congruence

Triangle Inequality Theorem

Triangle Inequality Theorem helps us decide whether three side lengths can actually make a triangle.

This idea is very important because not every set of three lengths can connect to form a closed shape. Sometimes the sides are too short or too long compared to each other, so they cannot meet.

In this lesson, you will learn what the Triangle Inequality Theorem says, how to test side lengths, and how to solve problems using it.

What is the Triangle Inequality Theorem?

The Triangle Inequality Theorem says that in any triangle, the sum of the lengths of any two sides must be greater than the length of the third side.

If a triangle has side lengths \(a\), \(b\), and \(c\), then all three of these must be true:

$$a+b>c$$

$$a+c>b$$

$$b+c>a$$

If even one of these is not true, then the three lengths cannot form a triangle.

Why does this make sense?

Imagine trying to connect two short sides to reach across a very long third side. If the two short sides together are not longer than the third side, they cannot bend enough to close the shape.

For example, if you had lengths 2, 3, and 10, the two smaller sides add to \(2+3=5\). Since 5 is less than 10, they cannot reach far enough to connect. So no triangle can be made.

An easy way to check

You can test all three inequalities, but there is also a shortcut.

If you put the side lengths in order from smallest to largest, you really only need to check whether the sum of the two smaller sides is greater than the largest side.

So if the sides are \(x\le y\le z\), check:

$$x+y>z$$

If this is true, the other two inequalities will also be true.

Important detail

The theorem says greater than, not greater than or equal to.

If the sum of two sides equals the third side, the sides form a straight line, not a triangle.

For example:

$$4+5=9$$

Since the sum is exactly equal to the third side, these lengths do not form a triangle.

Steps for checking side lengths

  1. Identify the three side lengths.

  2. Find the largest side.

  3. Add the two smaller sides.

  4. Compare that sum to the largest side.

  5. If the sum is greater, a triangle can be formed. If not, it cannot.

Worked Example 1: Can these lengths form a triangle?

Side lengths: 5, 7, 9

The largest side is 9. Add the two smaller sides:

$$5+7=12$$

Now compare:

$$12>9$$

Since the sum of the two smaller sides is greater than the largest side, these side lengths can form a triangle.

Worked Example 2: Can these lengths form a triangle?

Side lengths: 3, 4, 8

The largest side is 8. Add the two smaller sides:

$$3+4=7$$

Compare:

$$7<8$$

Since the sum of the two smaller sides is less than the largest side, these side lengths cannot form a triangle.

Worked Example 3: What if the sum is equal?

Side lengths: 6, 2, 4

The largest side is 6. Add the two smaller sides:

$$2+4=6$$

Compare:

$$6=6$$

This is not greater than 6. So these lengths do not form a triangle.

They would lie flat in a straight line instead of making a closed shape.

Worked Example 4: Find the possible value of a missing side

A triangle has side lengths 8, 11, and \(x\). What values can \(x\) have?

To form a triangle, the third side must be greater than the difference of the other two sides and less than their sum.

First find the difference:

$$11-8=3$$

Then find the sum:

$$11+8=19$$

So \(x\) must satisfy:

$$3<x<19$$

This means \(x\) must be greater than 3 and less than 19.

For example, 4, 10, and 18 would all work, but 3 and 19 would not.

Why does the missing side rule work?

If one side is unknown, it still has to fit with the Triangle Inequality Theorem.

For sides 8, 11, and \(x\), the inequalities are:

$$8+11>x$$

$$8+x>11$$

$$11+x>8$$

Simplifying gives:

$$19>x$$

$$x>3$$

$$x>-3$$

The condition \(x>-3\) is always true for a side length, since side lengths are positive. So the important result is:

$$3<x<19$$

Common mistakes to avoid

  • Using equals instead of greater than: If the sum equals the third side, it is not a triangle.

  • Checking only random pairs: Be sure you compare the two smaller sides to the largest side.

  • Forgetting which side is largest: Always identify the biggest number first.

  • Including impossible side lengths: A side length must be positive.

Quick practice ideas

Ask yourself these questions when you see three side lengths:

  • Which side is the largest?

  • Do the two smaller sides add to more than that largest side?

  • If there is a missing side, is it between the difference and the sum of the other two sides?

Summary

The Triangle Inequality Theorem tells us whether three side lengths can form a triangle.

For a triangle to exist, the sum of any two sides must be greater than the third side. In practice, this means the two smaller sides must add to more than the largest side.

If the sum is less than or equal to the largest side, no triangle can be formed. For a missing side, its length must be greater than the difference and less than the sum of the other two sides.

Put what you read to the test

You've worked through Triangle Inequality Theorem. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Triangle Angle Sum and Exterior Angle Theorems

Triangle Angle Sum and Exterior Angle Theorems

Triangles are one of the most important shapes in geometry. They may look simple, but they follow rules that are always true. In this lesson, you will learn two very useful triangle rules: the Triangle Angle Sum Theorem and the Exterior Angle Theorem.

These theorems help you find missing angle measures and explain how the angles in a triangle are connected. Once you understand these ideas, many triangle problems become much easier.

1. The Triangle Angle Sum Theorem

The Triangle Angle Sum Theorem says that the three interior angles of any triangle always add up to \(180^ ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{.} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}

If a triangle has angles \(A ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}, B, ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}and \(C ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{, then} $$m\angle A + m\angle B + m\angle C = 180^\circ$$

This rule is true for every triangle:

  • acute triangles
  • right triangles
  • obtuse triangles
  • scalene, isosceles, and equilateral triangles

So no matter what kind of triangle you have, the inside angles always total \(180^\circ ext{.} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}

How to use the Triangle Angle Sum Theorem

If you know two angles in a triangle, you can find the third by subtracting their sum from \(180^\circ ext{.} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}

For example, if two angles are \(50^\circ ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} and ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}65^\circ ext{, then the third angle is} $$180^\circ - (50^\circ + 65^\circ) = 180^\circ - 115^\circ = 65^\circ$$

2. The Exterior Angle of a Triangle

An exterior angle is formed when one side of a triangle is extended. This creates an angle on the outside of the triangle.

Each exterior angle is next to an interior angle. These two angles form a linear pair, which means they add up to \(180^\circ ext{.} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}

So if an interior angle is \(70^\circ ext{,} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} the exterior angle next to it is $$180^\circ - 70^\circ = 110^\circ$$

3. The Exterior Angle Theorem

The Exterior Angle Theorem says that an exterior angle of a triangle is equal to the sum of the two remote interior angles.

The remote interior angles are the two interior angles that are not next to the exterior angle.

If triangle \(ABC ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} has an exterior angle at vertex ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}C, ext{ then} $$m\text{ exterior angle at }C = m\angle A + m\angle B$$

This theorem is very helpful because it gives you another way to find missing angles.

Why the Exterior Angle Theorem makes sense

Suppose the interior angles of a triangle are \(A ext{, }B ext{, and }C ext{.} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}

From the Triangle Angle Sum Theorem, $$A + B + C = 180^\circ$$

If you extend one side at angle \(C ext{,} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} the exterior angle and ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}C form a straight line, so they add to ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}180^\circ ext{.} $$m\text{ exterior angle} + C = 180^\circ$$

Since both expressions equal \(180^\circ ext{,} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} the exterior angle must equal ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}A + B. $$m\text{ exterior angle} = A + B$$

4. Important ideas to remember

  • Interior angles are the angles inside the triangle.
  • Exterior angles are formed outside the triangle by extending a side.
  • The three interior angles of a triangle always total \(180^\circ ext{.}
  • An exterior angle equals the sum of the two remote interior angles.
  • An exterior angle and its adjacent interior angle form a straight line, so they add to \(180^\circ ext{.}

Worked Example 1: Find a missing interior angle

In a triangle, two angles measure \(48^\circ ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} and ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}73^\circ ext{. Find the third angle.}

Use the Triangle Angle Sum Theorem:

$$x + 48^\circ + 73^\circ = 180^\circ$$

Add the known angles:

$$48^\circ + 73^\circ = 121^\circ$$

Now subtract from \(180^\circ ext{:}

$$x = 180^\circ - 121^\circ = 59^\circ$$

Answer: The third angle is \(59^\circ ext{.}

Worked Example 2: Solve for a variable

A triangle has angle measures \(x ext{, }2x ext{, and }3x ext{. Find all three angles.}

Use the angle sum theorem:

$$x + 2x + 3x = 180^\circ$$

Combine like terms:

$$6x = 180^\circ$$

Solve for \(x ext{:}

$$x = 30^\circ$$

Now find each angle:

  • \(x = 30^\circ ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}
  • \(2x = 60^\circ ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}
  • \(3x = 90^\circ ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}

Answer: The angles are \(30^\circ ext{, }60^\circ ext{, and }90^\circ ext{.}

Worked Example 3: Use the Exterior Angle Theorem

A triangle has two remote interior angles measuring \(35^\circ ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} and ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}62^\circ ext{. Find the exterior angle.}

By the Exterior Angle Theorem:

$$m\text{ exterior angle} = 35^\circ + 62^\circ$$ $$m\text{ exterior angle} = 97^\circ$$

Answer: The exterior angle measures \(97^\circ ext{.}

Worked Example 4: Combine both theorems

An exterior angle of a triangle measures \(124^\circ ext{. One remote interior angle is }51^\circ ext{. Find the other remote interior angle and the adjacent interior angle.}

Let the other remote interior angle be \(x ext{.} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}

Use the Exterior Angle Theorem:

$$124^\circ = 51^\circ + x$$

Solve for \(x ext{:}

$$x = 124^\circ - 51^\circ = 73^\circ$$

Now find the adjacent interior angle. It forms a straight line with the exterior angle, so the two angles add to \(180^\circ ext{.}

$$\text{adjacent interior angle} = 180^\circ - 124^\circ = 56^\circ$$

Answer: The other remote interior angle is \(73^\circ ext{,} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} and the adjacent interior angle is ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}56^\circ ext{.}

Common mistakes to avoid

  • Do not forget all three interior angles must add to \(180^\circ ext{.}
  • Do not confuse the adjacent interior angle with a remote interior angle. The remote interior angles are the two angles far from the exterior angle.
  • Do not add the exterior angle to all three interior angles. The exterior angle equals only the sum of the two remote interior angles.
  • Check your arithmetic carefully. Many triangle problems are simple, but small addition or subtraction errors can lead to the wrong answer.

Quick problem-solving steps

  1. Decide whether you are working with interior angles or an exterior angle.
  2. If it is only interior angles, use $$m\angle A + m\angle B + m\angle C = 180^\circ$$
  3. If there is an exterior angle, use $$m\text{ exterior angle} = \text{sum of the two remote interior angles}$$
  4. If needed, also use the straight-line idea: $$\text{exterior angle} + \text{adjacent interior angle} = 180^\circ$$
  5. Solve carefully and check that your answer makes sense.

Summary

The Triangle Angle Sum Theorem tells us that the three interior angles of any triangle always add up to \(180^\circ ext{.} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{} ext{}

The Exterior Angle Theorem tells us that an exterior angle of a triangle equals the sum of the two remote interior angles. Also, an exterior angle and its adjacent interior angle form a straight line, so they add to \(180^\circ ext{.}

These two theorems are powerful tools for finding missing angles and understanding how triangle angles are related. With practice, you will be able to spot which rule to use and solve problems with confidence.

Put what you read to the test

You've worked through Triangle Angle Sum and Exterior Angle Theorems. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Isosceles and Equilateral Properties

Isosceles and Equilateral Properties

Triangles have many useful patterns. In this lesson, we will focus on two special kinds of triangles: isosceles triangles and equilateral triangles.

These triangles are important because their sides and angles are connected. If you know something about the side lengths, you can often figure out something about the angles. Also, if you know something about the angles, you can often figure out something about the side lengths.

This idea is called a bidirectional relationship. It means the rule works in both directions.

1. Key definitions

  • Isosceles triangle: a triangle with at least two congruent sides.
  • Equilateral triangle: a triangle with three congruent sides.
  • Congruent sides: sides that have the same length.
  • Congruent angles: angles that have the same measure.

In an isosceles triangle, the two equal sides are called the legs. The third side is called the base. The two angles opposite the equal sides are called the base angles.

2. Main property of isosceles triangles

The most important fact about an isosceles triangle is:

If two sides of a triangle are congruent, then the angles opposite those sides are congruent.

This is often called the Isosceles Triangle Theorem.

Suppose triangle \(ABC\) has \(AB = AC\). Since the equal sides are \(AB\) and \(AC\), the angles opposite them are \(\angle C\) and \(\angle B\). So:

$$AB = AC \implies \angle B = \angle C$$

This means the base angles of an isosceles triangle are equal.

3. The converse is also true

There is another very important rule that goes the other way:

If two angles of a triangle are congruent, then the sides opposite those angles are congruent.

This is called the Converse of the Isosceles Triangle Theorem.

In triangle \(ABC\), if \(\angle B = \angle C\), then the opposite sides are \(AC\) and \(AB\). So:

$$\angle B = \angle C \implies AB = AC$$

This is the bidirectional relationship for isosceles triangles:

  • Equal sides give equal opposite angles.
  • Equal angles give equal opposite sides.

4. Equilateral triangles

An equilateral triangle has three congruent sides. Since all three sides are equal, the angles opposite those sides must also be equal.

So in an equilateral triangle, all three angles are congruent.

Because the angle sum of any triangle is \(180^\circ\), each angle in an equilateral triangle measures:

$$\frac{180^\circ}{3} = 60^\circ$$

So every equilateral triangle is also equiangular, which means all angles are equal.

Important fact:

$$\text{Equilateral} \implies \text{all angles are } 60^\circ$$

The reverse is also true for triangles:

If a triangle has three congruent angles, then it has three congruent sides. So a triangle with all angles equal is equilateral.

5. Comparing isosceles and equilateral triangles

  • An isosceles triangle has at least two equal sides.
  • An equilateral triangle has exactly three equal sides.
  • Every equilateral triangle is also isosceles, because it has at least two equal sides.
  • Not every isosceles triangle is equilateral.

6. Using angle sum with isosceles triangles

Remember that the angles in any triangle add to \(180^\circ\):

$$\angle A + \angle B + \angle C = 180^\circ$$

If a triangle is isosceles, the base angles are equal. That makes it easier to find missing angles.

For example, if the two base angles are each \(x\) and the top angle is \(40^\circ\), then:

$$x + x + 40 = 180$$ $$2x + 40 = 180$$ $$2x = 140$$ $$x = 70$$

So each base angle is \(70^\circ\).

7. Worked Example 1: Find a missing angle in an isosceles triangle

Triangle \(PQR\) is isosceles with \(PQ = PR\). If \(\angle P = 44^\circ\), find \(\angle Q\) and \(\angle R\).

Step 1: Identify the equal sides.

Since \(PQ = PR\), the angles opposite those sides are equal. The angle opposite \(PQ\) is \(\angle R\), and the angle opposite \(PR\) is \(\angle Q\). So:

$$\angle Q = \angle R$$

Step 2: Use the triangle angle sum.

$$\angle P + \angle Q + \angle R = 180^\circ$$ $$44 + Q + R = 180$$

Since \(Q = R\), let each equal angle be \(x\):

$$44 + x + x = 180$$ $$44 + 2x = 180$$ $$2x = 136$$ $$x = 68$$

So:

$$\angle Q = 68^\circ, \quad \angle R = 68^\circ$$

Answer: \(\angle Q = 68^\circ\) and \(\angle R = 68^\circ\).

8. Worked Example 2: Use equal angles to find equal sides

In triangle \(ABC\), suppose \(\angle B = \angle C\). If \(AC = 9\), find \(AB\).

Step 1: Use the converse of the isosceles triangle theorem.

If \(\angle B = \angle C\), then the opposite sides are equal.

The side opposite \(\angle B\) is \(AC\). The side opposite \(\angle C\) is \(AB\). Therefore:

$$AC = AB$$

Step 2: Substitute the known value.

$$AC = 9 \implies AB = 9$$

Answer: \(AB = 9\).

9. Worked Example 3: Equilateral triangle angle measures

Triangle \(XYZ\) is equilateral. Find the measure of each angle.

Step 1: Use the equilateral triangle property.

All sides are congruent, so all angles are congruent.

Step 2: Use the angle sum of a triangle.

$$\angle X + \angle Y + \angle Z = 180^\circ$$

Since all three angles are equal, let each angle be \(x\):

$$x + x + x = 180$$ $$3x = 180$$ $$x = 60$$

Answer:

$$\angle X = \angle Y = \angle Z = 60^\circ$$

10. Worked Example 4: Decide what kind of triangle it is

A triangle has angles \(70^\circ\), \(70^\circ\), and \(40^\circ\). What can you conclude about the triangle?

Step 1: Look for equal angles.

Two angles are equal: \(70^\circ\) and \(70^\circ\).

Step 2: Use the converse of the isosceles triangle theorem.

If two angles are congruent, then the opposite sides are congruent.

So the triangle has two equal sides.

Conclusion: The triangle is isosceles.

It is not equilateral because not all three angles are equal.

11. Common mistakes to avoid

  • Mixing up opposite sides and opposite angles: Always match an angle with the side across from it.
  • Assuming every isosceles triangle is equilateral: Isosceles means at least two equal sides, not always three.
  • Forgetting the angle sum: The three angles in any triangle must add to \(180^\circ\).
  • Using the theorem in the wrong direction without checking: Remember there is a theorem and also a converse.

12. Quick checklist

When solving a problem about isosceles or equilateral triangles, ask yourself:

  1. Do I know any sides are congruent?
  2. If so, which angles are opposite those sides?
  3. Do I know any angles are congruent?
  4. If so, which sides are opposite those angles?
  5. Can I use the fact that triangle angles add to \(180^\circ\)?
  6. If the triangle is equilateral, can I use \(60^\circ\) for each angle?

13. Brief summary

In an isosceles triangle, if two sides are congruent, then the opposite angles are congruent. The converse is also true: if two angles are congruent, then the opposite sides are congruent.

In an equilateral triangle, all three sides are congruent, so all three angles are congruent. Since triangle angles add to \(180^\circ\), each angle in an equilateral triangle is \(60^\circ\).

These properties help you find missing sides, find missing angles, and classify triangles correctly.

Put what you read to the test

You've worked through Isosceles and Equilateral Properties. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Centers of a Triangle

Centers of a Triangle

Triangles have special points inside or around them called centers. These points are found by drawing certain lines, such as angle bisectors, perpendicular bisectors, medians, and altitudes.

In this lesson, you will learn the four most important triangle centers:

  • Incenter
  • Circumcenter
  • Centroid
  • Orthocenter

Each center is created in a different way and has a different meaning. Knowing these centers helps you understand the structure of a triangle and solve geometry problems.

Important idea: The word concurrent means that three or more lines meet at one point. In triangles, the special lines used to create the centers are concurrent.

1. The Incenter

The incenter is the point where the three angle bisectors of a triangle meet.

An angle bisector is a line or segment that divides an angle into two equal angles.

Properties of the incenter:

  • It is always inside the triangle.
  • It is the same distance from all three sides of the triangle.
  • It is the center of the triangle’s inscribed circle, also called the incircle.

This means that if you draw a circle centered at the incenter, and the circle just touches all three sides, that circle will fit perfectly inside the triangle.

How to construct the incenter:

  1. Draw a triangle.
  2. Construct the bisector of one angle.
  3. Construct the bisector of another angle.
  4. The point where they intersect is the incenter.

You only need two angle bisectors because the third will also pass through the same point.

2. The Circumcenter

The circumcenter is the point where the three perpendicular bisectors of the sides of a triangle meet.

A perpendicular bisector of a side is a line that:

  • passes through the midpoint of the side, and
  • forms a right angle with the side.

Properties of the circumcenter:

  • It is the same distance from all three vertices of the triangle.
  • It is the center of the triangle’s circumscribed circle, also called the circumcircle.
  • Its location depends on the type of triangle:
    • For an acute triangle, it is inside the triangle.
    • For a right triangle, it is at the midpoint of the hypotenuse.
    • For an obtuse triangle, it is outside the triangle.

How to construct the circumcenter:

  1. Draw a triangle.
  2. Find the perpendicular bisector of one side.
  3. Find the perpendicular bisector of another side.
  4. Their intersection point is the circumcenter.

Again, the third perpendicular bisector will pass through the same point.

3. The Centroid

The centroid is the point where the three medians of a triangle meet.

A median is a segment drawn from a vertex to the midpoint of the opposite side.

Properties of the centroid:

  • It is always inside the triangle.
  • It is often called the triangle’s balance point.
  • It divides each median in a 2:1 ratio.

The 2:1 ratio means the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint of the opposite side.

If the full median has length 12, then:

$$ \text{vertex to centroid} = 8 $$ $$ \text{centroid to midpoint} = 4 $$

How to construct the centroid:

  1. Draw a triangle.
  2. Find the midpoint of one side.
  3. Draw a segment from the opposite vertex to that midpoint.
  4. Do the same for another side.
  5. The point where the medians meet is the centroid.

4. The Orthocenter

The orthocenter is the point where the three altitudes of a triangle meet.

An altitude is a segment drawn from a vertex perpendicular to the line containing the opposite side.

Properties of the orthocenter:

  • For an acute triangle, it is inside the triangle.
  • For a right triangle, it is at the right-angle vertex.
  • For an obtuse triangle, it is outside the triangle.

How to construct the orthocenter:

  1. Draw a triangle.
  2. From one vertex, draw a line perpendicular to the opposite side.
  3. From another vertex, draw another perpendicular line to the opposite side.
  4. The point where the altitudes intersect is the orthocenter.

Sometimes you need to extend a side of the triangle to draw an altitude, especially in an obtuse triangle.

Comparing the Four Centers

Center Formed by Special property Usual location
Incenter Angle bisectors Equal distance from sides Always inside
Circumcenter Perpendicular bisectors Equal distance from vertices Inside, on, or outside
Centroid Medians Balance point; 2:1 ratio Always inside
Orthocenter Altitudes Intersection of perpendiculars from vertices Inside, on, or outside

Helpful way to remember:

  • Incenter incircle  inside the triangle
  • Circumcenter circumcircle around the triangle
  • Centroid center of mass or balance
  • Orthocenter  formed by altitudes at right angles

Worked Example 1: Identifying a center from its construction

Suppose the three angle bisectors of a triangle meet at point \(P\). What center is \(P\)?

Step 1: Look at the type of lines used.

The lines are angle bisectors.

Step 2: Match this with the correct center.

The point where the angle bisectors meet is the incenter.

Answer: \(P\) is the incenter.

Worked Example 2: Using the centroid’s 2:1 ratio

In triangle \(ABC\), median \(AD\) has length \(15\). Point \(G\) is the centroid. Find \(AG\) and \(GD\).

Step 1: Recall the centroid divides a median in a \(2:1\) ratio.

This means:

$$ AG:GD = 2:1 $$

Step 2: Add the ratio parts.

$$ 2+1=3 $$

So the full median is divided into 3 equal parts.

Step 3: Find one part.

$$ 15 \div 3 = 5 $$

Step 4: Find each segment.

$$ AG = 2 \times 5 = 10 $$ $$ GD = 1 \times 5 = 5 $$

Answer: \(AG=10\) and \(GD=5\).

Worked Example 3: Locating the circumcenter

A triangle is a right triangle. Where is its circumcenter located?

Step 1: Recall the circumcenter is where the perpendicular bisectors meet.

Step 2: Use the special rule for right triangles.

In a right triangle, the circumcenter lies at the midpoint of the hypotenuse.

Answer: The circumcenter is at the midpoint of the hypotenuse.

Worked Example 4: Identifying which center can be outside

Which triangle centers can lie outside a triangle?

Step 1: Check each center.

  • The incenter is always inside.
  • The centroid is always inside.
  • The circumcenter can be outside in an obtuse triangle.
  • The orthocenter can be outside in an obtuse triangle.

Answer: The circumcenter and orthocenter can lie outside a triangle.

Common Mistakes to Avoid

  • Do not confuse an angle bisector with a median.
  • Do not confuse a perpendicular bisector with an altitude.
  • Remember: the incenter is equal distance from sides, not vertices.
  • Remember: the circumcenter is equal distance from vertices, not sides.
  • For the centroid, the longer part of the median is from the vertex to the centroid.

Quick Check

  1. What lines form the incenter?
  2. Which center is the balance point of a triangle?
  3. Which center is the same distance from all three vertices?
  4. Where is the orthocenter in a right triangle?

Answers:

  1. Angle bisectors
  2. Centroid
  3. Circumcenter
  4. At the right-angle vertex

Summary

A triangle has four major centers, and each comes from a different set of concurrent lines.

  • The incenter is formed by angle bisectors and is the center of the incircle.
  • The circumcenter is formed by perpendicular bisectors and is the center of the circumcircle.
  • The centroid is formed by medians and divides each median in a \(2:1\) ratio.
  • The orthocenter is formed by altitudes.

If you can identify the type of lines being used, you can identify the triangle center. That is the key idea of this topic.

Put what you read to the test

You've worked through Centers of a Triangle. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Congruence Transformations

Congruence Transformations are movements that change the position of a figure without changing its size or shape.

In geometry, these movements are called rigid transformations. The word rigid means “stiff,” so the figure does not stretch, shrink, or bend. If one figure can be moved onto another using rigid transformations, then the figures are congruent.

This idea is very important with triangles. If you can show that one triangle can be translated, reflected, or rotated to match another triangle exactly, then the triangles are congruent.

In this lesson, you will learn:

  • what congruence means,
  • the three main congruence transformations,
  • how to describe a sequence of transformations,
  • and how to use transformations to show that two figures are congruent.

1. What does congruent mean?

Two figures are congruent if they have exactly the same size and shape.

This means:

  • all corresponding side lengths are equal, and
  • all corresponding angle measures are equal.

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

$$\triangle ABC \cong \triangle DEF$$

The order of the letters matters. It tells you which vertices match:

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

2. The three congruence transformations

There are three main rigid transformations:

  1. Translation
  2. Reflection
  3. Rotation

Each of these keeps lengths and angles the same.

A. Translation

A translation slides a figure from one place to another.

Every point moves the same distance in the same direction.

For example, if a point \((x,y)\) is translated 4 units right and 2 units up, its image becomes

$$ (x,y) \to (x+4, y+2) $$

Important facts about translations:

  • The figure does not turn.
  • The figure does not flip.
  • Side lengths and angle measures stay the same.

B. Reflection

A reflection flips a figure across a line called the line of reflection.

The reflected figure is like a mirror image.

Common reflections on a coordinate plane:

  • Across the \(x\)-axis: \((x,y) \to (x,-y)\)
  • Across the \(y\)-axis: \((x,y) \to (-x,y)\)
  • Across the line \(y=x\): \((x,y) \to (y,x)\)

Important facts about reflections:

  • The figure keeps the same size and shape.
  • The orientation is reversed, like looking in a mirror.

C. Rotation

A rotation turns a figure around a fixed point called the center of rotation.

The turn can be clockwise or counterclockwise.

Common rotations about the origin:

  • \(90^\circ\) counterclockwise: \((x,y) \to (-y,x)\)
  • \(180^\circ\): \((x,y) \to (-x,-y)\)
  • \(270^\circ\) counterclockwise: \((x,y) \to (y,-x)\)

Important facts about rotations:

  • The figure turns, but does not change size or shape.
  • Distances from the center of rotation stay the same.

3. Why these transformations prove congruence

Translations, reflections, and rotations are all rigid transformations. Since they do not change lengths or angles, the image has the same size and shape as the original figure.

So, if you can move one figure onto another exactly using one or more of these transformations, then the two figures are congruent.

You can think of it this way: if one figure fits perfectly on top of the other after sliding, flipping, or turning, then they are congruent.

4. Sequence of transformations

Sometimes one transformation is not enough. You may need a sequence of transformations, which means doing more than one movement in order.

For example:

  • translate, then reflect
  • rotate, then translate
  • reflect, then rotate

The order matters. Doing a reflection first and then a translation may give a different result than translating first and then reflecting.

When describing a sequence, be specific:

  • what transformation is used,
  • how far or how much the figure moves,
  • and, if needed, the line of reflection or center of rotation.

5. How to tell which transformation may be needed

Here are some clues:

  • If the figure has the same orientation and is just in a different place, think translation.
  • If the figure looks like a mirror image, think reflection.
  • If the figure looks turned, think rotation.

Sometimes more than one sequence can work. In geometry, you only need to show one correct sequence that maps one figure onto the other.

6. Worked Examples

Example 1: Translation of a triangle

Triangle \(ABC\) has vertices \(A(1,2)\), \(B(3,2)\), and \(C(2,5)\).

Triangle \(A'B'C'\) has vertices \(A'(5,1)\), \(B'(7,1)\), and \(C'(6,4)\).

Show that the triangles are congruent using a transformation.

Step 1: Compare each point to its image.

  • \(A(1,2) \to A'(5,1)\)
  • \(B(3,2) \to B'(7,1)\)
  • \(C(2,5) \to C'(6,4)\)

Each point moves 4 units right and 1 unit down.

Step 2: Write the translation rule.

$$ (x,y) \to (x+4, y-1) $$

Step 3: State the conclusion.

A translation maps \(\triangle ABC\) onto \(\triangle A'B'C'\). Since a translation is a rigid transformation,

$$ \triangle ABC \cong \triangle A'B'C' $$

Example 2: Reflection across the \(y\)-axis

Triangle \(PQR\) has vertices \(P(2,1)\), \(Q(5,1)\), and \(R(3,4)\).

Triangle \(P'Q'R'\) has vertices \(P'(-2,1)\), \(Q'(-5,1)\), and \(R'(-3,4)\).

Determine the transformation and explain why the triangles are congruent.

Step 1: Look for a pattern.

  • \((2,1) \to (-2,1)\)
  • \((5,1) \to (-5,1)\)
  • \((3,4) \to (-3,4)\)

The \(x\)-values change sign, but the \(y\)-values stay the same.

Step 2: Identify the rule.

$$ (x,y) \to (-x,y) $$

This is a reflection across the \(y\)-axis.

Step 3: State the conclusion.

A reflection across the \(y\)-axis maps \(\triangle PQR\) onto \(\triangle P'Q'R'\). Therefore,

$$ \triangle PQR \cong \triangle P'Q'R' $$

Example 3: Rotation about the origin

Triangle \(LMN\) has vertices \(L(1,2)\), \(M(4,2)\), and \(N(2,5)\).

Triangle \(L'M'N'\) has vertices \(L'(-2,1)\), \(M'(-2,4)\), and \(N'(-5,2)\).

Show that one triangle is a rotation of the other.

Step 1: Test a common rotation rule.

For a \(90^\circ\) counterclockwise rotation about the origin:

$$ (x,y) \to (-y,x) $$

Apply it:

  • \(L(1,2) \to (-2,1)\)
  • \(M(4,2) \to (-2,4)\)
  • \(N(2,5) \to (-5,2)\)

The image points match exactly.

Step 2: State the conclusion.

A \(90^\circ\) counterclockwise rotation about the origin maps \(\triangle LMN\) onto \(\triangle L'M'N'\). So,

$$ \triangle LMN \cong \triangle L'M'N' $$

Example 4: A sequence of transformations

Suppose triangle \(ABC\) is reflected across the \(x\)-axis and then translated 3 units right.

The original points are \(A(1,2)\), \(B(4,2)\), and \(C(2,5)\).

Find the final image of the triangle.

Step 1: Reflect across the \(x\)-axis.

Rule:

$$ (x,y) \to (x,-y) $$

  • \(A(1,2) \to (1,-2)\)
  • \(B(4,2) \to (4,-2)\)
  • \(C(2,5) \to (2,-5)\)

Step 2: Translate 3 units right.

Rule:

$$ (x,y) \to (x+3,y) $$

  • \((1,-2) \to (4,-2)\)
  • \((4,-2) \to (7,-2)\)
  • \((2,-5) \to (5,-5)\)

Final image:

  • \(A'(4,-2)\)
  • \(B'(7,-2)\)
  • \(C'(5,-5)\)

This sequence uses two rigid transformations, so the final image is still congruent to the original triangle.

7. Connecting transformations to triangle congruence

When two triangles are congruent, one can be mapped onto the other by a sequence of rigid transformations.

This means:

  • corresponding sides match in length,
  • corresponding angles match in measure,
  • and one triangle fits exactly on the other after the movements.

For example, if triangle \(ABC\) can be translated and then rotated so that it lands exactly on triangle \(DEF\), then

$$ \triangle ABC \cong \triangle DEF $$

8. Common mistakes to avoid

  • Mixing up congruent and similar: Congruent figures have the same size and shape. Similar figures have the same shape but not always the same size.
  • Ignoring order of vertices: In a congruence statement, the matching vertices must be written in the correct order.
  • Using a non-rigid transformation: Stretching or shrinking is not allowed for congruence transformations.
  • Forgetting orientation changes: A reflection flips a figure, while a translation does not.
  • Doing transformations in the wrong order: In a sequence, order matters.

9. Quick checklist for solving problems

  1. Look at the original figure and the image.
  2. Decide whether it was slid, flipped, turned, or moved by more than one step.
  3. Check corresponding points.
  4. Write the transformation or sequence clearly.
  5. Conclude that the figures are congruent because rigid transformations preserve size and shape.

Summary

Congruence transformations are translations, reflections, and rotations. These are rigid transformations, so they keep side lengths and angle measures the same.

If one figure can be mapped onto another by a sequence of these transformations, then the figures are congruent. In triangle geometry, this gives a powerful way to prove that two triangles are exactly the same size and shape.

Put what you read to the test

You've worked through Congruence Transformations. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Triangle Congruence Postulates

Triangle Congruence Postulates help us decide when two triangles are exactly the same size and shape. When two triangles are congruent, every matching side has the same length and every matching angle has the same measure.

We write triangle congruence using the symbol \(\cong\). For example, if triangle \(ABC\) is congruent to triangle \(DEF\), we write \(\triangle ABC \cong \triangle DEF\).

This does not mean the triangles must be facing the same direction. One triangle can be turned, flipped, or moved and still be congruent to the other.

In this lesson, you will learn the main ways to prove triangles are congruent:

  • SSS: Side-Side-Side
  • SAS: Side-Angle-Side
  • ASA: Angle-Side-Angle
  • AAS: Angle-Angle-Side
  • HL: Hypotenuse-Leg for right triangles

These are called congruence postulates or criteria. They let us prove triangles are congruent without checking all 6 parts.

Why does this matter? If you know two triangles are congruent, then all their matching parts are equal. This idea is often summarized as CPCTC, which means Corresponding Parts of Congruent Triangles are Congruent.

Before using any postulate, it is very important to match the triangles correctly. The order of the letters tells which parts correspond.

For example, if

$$\triangle ABC \cong \triangle DEF,$$

then the matching parts are:

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

So the corresponding sides are:

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

And the corresponding angles are:

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

Now let’s look at each congruence postulate.

1. SSS: Side-Side-Side

If all three sides of one triangle are congruent to all three matching sides of another triangle, then the triangles are congruent.

In symbols, if

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

then

$$\triangle ABC \cong \triangle DEF \text{ by SSS.}$$

SSS works because three side lengths completely determine the shape of a triangle.

2. SAS: Side-Angle-Side

If two sides and the included angle between them in one triangle are congruent to two sides and the included angle in another triangle, then the triangles are congruent.

The word included is very important. It means the angle must be between the two given sides.

For example, if

$$AB \cong DE, \quad \angle B \cong \angle E, \quad BC \cong EF,$$

then

$$\triangle ABC \cong \triangle DEF \text{ by SAS.}$$

3. ASA: Angle-Side-Angle

If two angles and the included side between them in one triangle are congruent to two angles and the included side in another triangle, then the triangles are congruent.

Example:

$$\angle A \cong \angle D, \quad AB \cong DE, \quad \angle B \cong \angle E$$

gives

$$\triangle ABC \cong \triangle DEF \text{ by ASA.}$$

Here, the side is between the two known angles.

4. AAS: Angle-Angle-Side

If two angles and a non-included side in one triangle are congruent to two angles and the matching non-included side in another triangle, then the triangles are congruent.

Example:

$$\angle A \cong \angle D, \quad \angle B \cong \angle E, \quad AC \cong DF$$

gives

$$\triangle ABC \cong \triangle DEF \text{ by AAS.}$$

AAS works because if two angles are known, the third angle is automatically fixed since the angles in a triangle add up to \(180^\circ\).

5. HL: Hypotenuse-Leg

HL is only for right triangles. If the hypotenuse and one leg of one right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent.

Remember:

  • The hypotenuse is the side opposite the right angle.
  • A leg is one of the two sides that form the right angle.

If \(\triangle ABC\) and \(\triangle DEF\) are right triangles, and

$$AC \cong DF \quad \text{(hypotenuse)}$$

and

$$AB \cong DE \quad \text{(a leg)},$$

then

$$\triangle ABC \cong \triangle DEF \text{ by HL.}$$

What does not work?

Not every group of matching parts proves congruence.

  • AAA does not prove congruence. It only proves the triangles are similar, meaning same shape but not necessarily same size.
  • SSA does not usually prove congruence. This is not a valid general congruence postulate.

So when proving triangles congruent, be careful to use only SSS, SAS, ASA, AAS, or HL.

How to choose the correct postulate

  1. Look at the information given.
  2. Decide whether the given parts are sides, angles, or both.
  3. Check whether the angle or side is included.
  4. If the triangles are right triangles, see if HL applies.
  5. Name the triangles in corresponding order.

Worked Example 1: Using SSS

Suppose you are given:

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

Can you prove \(\triangle ABC \cong \triangle DEF\)?

Step 1: Count the matching parts. We have three pairs of congruent sides.

Step 2: This matches the SSS postulate.

Conclusion:

$$\triangle ABC \cong \triangle DEF \text{ by SSS.}$$

Worked Example 2: Using SAS

Given:

  • \(AB = PQ\)
  • \(\angle B = \angle Q\)
  • \(BC = QR\)

Prove \(\triangle ABC \cong \triangle PQR\).

Step 1: We have two pairs of sides: \(AB = PQ\) and \(BC = QR\).

Step 2: We also have one pair of angles: \(\angle B = \angle Q\).

Step 3: The angle is between the two known sides in each triangle. So it is the included angle.

Conclusion:

$$\triangle ABC \cong \triangle PQR \text{ by SAS.}$$

Worked Example 3: Deciding between ASA and AAS

Given:

  • \(\angle M = \angle X\)
  • \(\angle N = \angle Y\)
  • \(MP = XZ\)

Can we prove \(\triangle MNP \cong \triangle XYZ\)?

Step 1: We have two pairs of angles.

Step 2: We have one pair of sides: \(MP = XZ\).

Step 3: Ask whether the given side is between the two given angles.

The side between \(\angle M\) and \(\angle N\) would be \(MN\), not \(MP\). So the side is not included.

Conclusion:

$$\triangle MNP \cong \triangle XYZ \text{ by AAS.}$$

Worked Example 4: Using HL in right triangles

Suppose \(\triangle JKL\) and \(\triangle RST\) are right triangles with right angles at \(K\) and \(S\).

Given:

  • \(JL = RT\)
  • \(JK = RS\)

Can we prove the triangles are congruent?

Step 1: Since both triangles are right triangles, HL might apply.

Step 2: In \(\triangle JKL\), the side opposite the right angle is \(JL\), so it is the hypotenuse.

In \(\triangle RST\), the side opposite the right angle is \(RT\), so it is the hypotenuse.

Step 3: We are also given one pair of corresponding legs: \(JK = RS\).

Conclusion:

$$\triangle JKL \cong \triangle RST \text{ by HL.}$$

Important reminder about naming triangles

When writing a congruence statement, the order must match corresponding parts.

For example, if:

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

then you should write:

$$\triangle ABC \cong \triangle DEF$$

and not something like \(\triangle ABC \cong \triangle DFE\), because that would mismatch the parts.

Quick checklist

  • Do I have 3 sides? Use SSS.
  • Do I have 2 sides and the included angle? Use SAS.
  • Do I have 2 angles and the included side? Use ASA.
  • Do I have 2 angles and a non-included side? Use AAS.
  • Are they right triangles with a hypotenuse and one leg? Use HL.
  • Do I only have AAA or SSA? That is not enough to prove congruence.

Brief Summary

Triangle congruence means two triangles are exactly the same size and shape. You can prove triangles congruent using SSS, SAS, ASA, AAS, and HL. The key is to look carefully at the given information, decide which pattern it matches, and write the triangles in the correct corresponding order.

Put what you read to the test

You've worked through Triangle Congruence Postulates. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

CPCTC Applications

CPCTC Applications

When two triangles are proven congruent, they have exactly the same size and shape. That means every matching side has the same length, and every matching angle has the same measure.

This idea is called CPCTC, which stands for Corresponding Parts of Congruent Triangles are Congruent.

CPCTC is very useful because once you prove two triangles are congruent, you can use their matching parts to find unknown side lengths or angle measures.

Important idea: CPCTC comes after you prove triangles congruent. You cannot use CPCTC first. First prove the triangles are congruent using a triangle congruence rule like:

  • SSS: 3 pairs of equal sides
  • SAS: 2 pairs of equal sides and the included angle
  • ASA: 2 pairs of equal angles and the included side
  • AAS: 2 pairs of equal angles and a non-included side
  • HL: for right triangles only, hypotenuse and one leg

What does “corresponding” mean?

Corresponding parts are the parts that match in the two triangles. For example, if:

$$\triangle ABC \cong \triangle DEF$$

then the order tells you which parts correspond:

  • Vertex \(A\) corresponds to vertex \(D\)

  • Vertex \(B\) corresponds to vertex \(E\)

  • Vertex \(C\) corresponds to vertex \(F\)

So the corresponding sides are:

  • \(AB \cong DE\)

  • \(BC \cong EF\)

  • \(AC \cong DF\)

And the corresponding angles are:

  • \(\angle A \cong \angle D\)

  • \(\angle B \cong \angle E\)

  • \(\angle C \cong \angle F\)

If you match the letters incorrectly, you may compare the wrong sides or angles. So always check the order carefully.

How to use CPCTC

  1. Identify the two triangles.

  2. Use given information to prove the triangles congruent.

  3. Match the corresponding vertices in the correct order.

  4. Use CPCTC to state that matching sides or angles are congruent.

  5. Solve for any unknown values.

Worked Example 1: Finding an unknown side

Suppose you know:

  • \(AB = DE\)

  • \(BC = EF\)

  • \(AC = DF\)

Then:

$$\triangle ABC \cong \triangle DEF \quad \text{by SSS}$$

If \(AB = 3x + 2\) and \(DE = 17\), find \(x\).

Since the triangles are congruent, corresponding sides are congruent. Here, \(AB\) corresponds to \(DE\), so:

$$AB = DE$$

Substitute the expressions:

$$3x + 2 = 17$$

Solve:

$$3x = 15$$

$$x = 5$$

So, the value of \(x\) is 5.

Worked Example 2: Finding an unknown angle

Given:

$$\triangle PQR \cong \triangle STU$$

This means:

  • \(P \leftrightarrow S\)

  • \(Q \leftrightarrow T\)

  • \(R \leftrightarrow U\)

If \(m\angle Q = 4x + 10\) and \(m\angle T = 46^\circ\), find \(x\).

Because corresponding angles in congruent triangles are congruent:

$$\angle Q \cong \angle T$$

So their measures are equal:

$$4x + 10 = 46$$

Solve:

$$4x = 36$$

$$x = 9$$

So, \(x = 9\).

Worked Example 3: Prove congruent first, then use CPCTC

Suppose in triangles \(\triangle MNO\) and \(\triangle PQR\), you know:

  • \(MN = PQ\)

  • \(NO = QR\)

  • \(\angle N = \angle Q\)

You want to show that \(MO = PR\).

First, look at the information. We have two sides and the included angle:

  • \(MN = PQ\)

  • \(NO = QR\)

  • \(\angle N = \angle Q\)

So:

$$\triangle MNO \cong \triangle PQR \quad \text{by SAS}$$

Now use CPCTC. The correspondence is:

  • \(M \leftrightarrow P\)

  • \(N \leftrightarrow Q\)

  • \(O \leftrightarrow R\)

Therefore, side \(MO\) corresponds to side \(PR\). By CPCTC:

$$MO = PR$$

This example shows the correct order:

  1. Prove congruent using SAS.

  2. Use CPCTC to conclude that corresponding sides are equal.

Worked Example 4: Using CPCTC in a diagram with a shared side

Consider triangles \(\triangle ABD\) and \(\triangle CBD\). Suppose you know:

  • \(AB = CB\)

  • \(\angle ADB = \angle CDB\)

  • \(BD\) is a shared side

Can we conclude that \(\triangle ABD \cong \triangle CBD\)? Not yet, because the given angle is not enough with the two sides we have listed. We need the side-angle-side pattern to match correctly.

Instead, suppose the information is:

  • \(AB = CB\)

  • \(BD = BD\)

  • \(\angle ABD = \angle DBC\)

Now we have two sides and the included angle, so:

$$\triangle ABD \cong \triangle CBD \quad \text{by SAS}$$

Then by CPCTC, the corresponding angles at \(A\) and \(C\) are congruent:

$$\angle A \cong \angle C$$

Also, the corresponding sides \(AD\) and \(CD\) are congruent:

$$AD = CD$$

If \(AD = 2x - 1\) and \(CD = 9\), then:

$$2x - 1 = 9$$

$$2x = 10$$

$$x = 5$$

Common mistakes to avoid

  • Using CPCTC before proving congruence. You must prove the triangles congruent first.

  • Matching the wrong vertices. Always use the order of the congruence statement.

  • Using the wrong congruence rule. Check whether the information fits SSS, SAS, ASA, AAS, or HL.

  • Forgetting shared sides. If two triangles share a side, that side is equal to itself.

Quick check

1. If \(\triangle XYZ \cong \triangle LMN\), which angle corresponds to \(\angle Z\)?

Answer: \(\angle N\)

2. If \(\triangle ABC \cong \triangle DEF\) and \(AC = 12\), what is \(DF\)?

Answer: \(12\), because \(AC\) corresponds to \(DF\).

3. If \(\triangle RST \cong \triangle UVW\) and \(m\angle S = 3x + 5\), \(m\angle V = 26^\circ\), find \(x\).

Since \(\angle S\) corresponds to \(\angle V\):

$$3x + 5 = 26$$

$$3x = 21$$

$$x = 7$$

Summary

CPCTC means Corresponding Parts of Congruent Triangles are Congruent. It lets you conclude that matching sides and angles are equal, but only after you prove the triangles are congruent.

To use CPCTC successfully, first prove triangle congruence, then match the correct vertices, and finally use those matching parts to solve for unknown side lengths or angle measures.

Put what you read to the test

You've worked through CPCTC Applications. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Coordinate Geometry Proofs for Triangles

Coordinate Geometry Proofs for Triangles

In geometry, we often prove facts about triangles by using diagrams, angle rules, and congruence. In coordinate geometry proofs, we place the triangle on a coordinate plane and use formulas to prove things are true.

This is a powerful method because coordinates let us turn a picture into numbers. Then we can use those numbers to show that sides are equal, slopes match, or midpoints line up exactly.

In this lesson, you will learn how to use three main tools:

  • Distance formula to compare side lengths
  • Midpoint formula to find the point halfway between two endpoints
  • Slope formula to show lines are parallel or perpendicular

These tools help you classify triangles and write coordinate proofs for triangle properties.

1. The Coordinate Plane Review

A point on the coordinate plane is written as \((x,y)\). The first number tells how far left or right the point is, and the second number tells how far up or down it is.

If a triangle has vertices \(A(x_1,y_1)\), \(B(x_2,y_2)\), and \(C(x_3,y_3)\), then each side of the triangle is just a line segment connecting two of those points.

2. Formulas You Need

Slope Formula

The slope of the line through \((x_1,y_1)\) and \((x_2,y_2)\) is

$$m=\frac{y_2-y_1}{x_2-x_1}$$

Slope tells how steep a line is.

  • If two lines have the same slope, they are parallel.
  • If their slopes are negative reciprocals, they are perpendicular.

For example, if one line has slope \(2\), a perpendicular line has slope \(-\frac{1}{2}\).

Distance Formula

The distance between \((x_1,y_1)\) and \((x_2,y_2)\) is

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

This formula comes from the Pythagorean Theorem. It helps us decide whether sides of a triangle are equal.

Midpoint Formula

The midpoint of a segment with endpoints \((x_1,y_1)\) and \((x_2,y_2)\) is

$$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$

The midpoint is the point exactly halfway between the endpoints.

3. How These Formulas Help Prove Triangle Types

In coordinate geometry, we often prove a triangle is a certain type by checking its side lengths or angle relationships.

  • Isosceles triangle: at least two equal sides
  • Equilateral triangle: all three sides equal
  • Right triangle: one right angle
  • Scalene triangle: no equal sides

To prove these, we usually do one of the following:

  • Use the distance formula to compare side lengths
  • Use the slope formula to show two sides are perpendicular, which proves a right angle
  • Use the midpoint formula to show a segment is being bisected

4. Writing a Coordinate Proof

A coordinate proof should be organized and logical. A good structure is:

  1. Write the coordinates of the points.
  2. State what you are trying to prove.
  3. Use the correct formula(s).
  4. Show your calculations clearly.
  5. State the conclusion based on the results.

For example, if two side lengths are equal, you can conclude the triangle is isosceles. If two slopes are negative reciprocals, you can conclude the angle is a right angle.

5. Important Patterns to Remember

  • If \(AB=AC\), then triangle \(ABC\) is isosceles.
  • If slopes of \(AB\) and \(AC\) are negative reciprocals, then \(\angle A\) is a right angle.
  • If the midpoint of one segment is also the midpoint of another, the segments share the same halfway point.
  • When comparing distances, it is often okay to compare the squares of distances to avoid square roots.

For example, instead of comparing \(\sqrt{13}\) and \(\sqrt{13}\), you can compare \(13\) and \(13\).

Worked Example 1: Proving a Triangle is Isosceles

Given \(A(1,2)\), \(B(5,4)\), and \(C(3,8)\), prove that triangle \(ABC\) is isosceles.

Step 1: Find \(AB\)

$$AB=\sqrt{(5-1)^2+(4-2)^2}=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20}$$

Step 2: Find \(AC\)

$$AC=\sqrt{(3-1)^2+(8-2)^2}=\sqrt{2^2+6^2}=\sqrt{4+36}=\sqrt{40}$$

Step 3: Find \(BC\)

$$BC=\sqrt{(3-5)^2+(8-4)^2}=\sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20}$$

Step 4: Compare side lengths

We found that \(AB=\sqrt{20}\) and \(BC=\sqrt{20}\).

Since two sides are equal, triangle \(ABC\) is isosceles.

Worked Example 2: Proving a Triangle is Right

Given \(A(0,0)\), \(B(4,0)\), and \(C(4,3)\), prove that triangle \(ABC\) is a right triangle.

We will use slope to check whether two sides are perpendicular.

Step 1: Find the slope of \(AB\)

$$m_{AB}=\frac{0-0}{4-0}=0$$

This means \(AB\) is a horizontal line.

Step 2: Find the slope of \(BC\)

$$m_{BC}=\frac{3-0}{4-4}=\frac{3}{0}$$

This slope is undefined, which means \(BC\) is a vertical line.

Step 3: Use the relationship

A horizontal line and a vertical line are perpendicular.

So \(AB \perp BC\), which means \(\angle B\) is a right angle.

Therefore, triangle \(ABC\) is a right triangle.

Worked Example 3: Proving a Triangle is Right Isosceles

Given \(A(0,0)\), \(B(2,2)\), and \(C(2,-2)\), classify triangle \(ABC\).

We will check both side lengths and slopes.

Step 1: Find \(AB\)

$$AB=\sqrt{(2-0)^2+(2-0)^2}=\sqrt{4+4}=\sqrt{8}$$

Step 2: Find \(AC\)

$$AC=\sqrt{(2-0)^2+(-2-0)^2}=\sqrt{4+4}=\sqrt{8}$$

Since \(AB=AC\), the triangle is isosceles.

Step 3: Check for a right angle using slopes

$$m_{AB}=\frac{2-0}{2-0}=1$$

$$m_{AC}=\frac{-2-0}{2-0}=-1$$

The slopes \(1\) and \(-1\) are negative reciprocals, so \(AB\) and \(AC\) are perpendicular.

This means \(\angle A\) is a right angle.

Conclusion: Triangle \(ABC\) is a right isosceles triangle.

Worked Example 4: Using Midpoint to Prove a Triangle Property

Suppose triangle \(ABC\) has points \(A(0,4)\), \(B(-2,0)\), and \(C(2,0)\). Show that point \(D(0,0)\) is the midpoint of \(BC\), and explain what this tells us about segment \(AD\).

Step 1: Find the midpoint of \(BC\)

Using \(B(-2,0)\) and \(C(2,0)\),

$$\left(\frac{-2+2}{2},\frac{0+0}{2}\right)=(0,0)$$

The midpoint of \(BC\) is \((0,0)\), which is point \(D\).

Step 2: State the result

Since \(D\) is the midpoint of \(BC\), segment \(AD\) goes from vertex \(A\) to the midpoint of the opposite side.

That means \(AD\) is a median of triangle \(ABC\).

This example shows how midpoint formulas can help prove triangle segments have special roles.

6. Comparing Distance Squared

Sometimes square roots make calculations look messy. If you only need to know whether lengths are equal, you can compare the values inside the square roots.

For example, suppose:

$$AB=\sqrt{29}, \qquad AC=\sqrt{29}$$

Since both are based on the same number, the lengths are equal.

This is especially helpful when proving triangles are isosceles.

7. Common Mistakes to Avoid

  • Mixing up the slope formula: Keep the order the same in the numerator and denominator.
  • Arithmetic errors: Be careful with negative numbers, especially when squaring.
  • Forgetting what the result means: Equal lengths mean isosceles; perpendicular sides mean a right triangle.
  • Using only the picture: In a proof, the calculations are what justify the conclusion.

8. Strategy Guide: Which Formula Should You Use?

  • If you need to show sides are equal, use the distance formula.
  • If you need to show an angle is \(90^\circ\), use slope.
  • If you need to show a point is halfway, use the midpoint formula.
  • If a problem asks you to classify a triangle fully, you may need more than one formula.

9. Final Summary

Coordinate geometry proofs let you use algebra to prove facts about triangles. The three key tools are slope, distance, and midpoint.

Use distance to compare side lengths, slope to check for parallel or perpendicular lines, and midpoint to show a segment is bisected. With these formulas, you can prove whether a triangle is isosceles, right, or has special segments like medians.

When solving, always show your work clearly and connect your calculations to your conclusion. That is what turns a calculation into a proof.

Put what you read to the test

You've worked through Coordinate Geometry Proofs for Triangles. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.