Chapter 10

Geometric Constructions

Bisectors and Basic Angle Constructions

Bisectors and Basic Angle Constructions

In geometry, a construction is a drawing made using only a compass and a straightedge. A straightedge helps you draw straight lines, but it does not measure. A compass helps you copy distances and draw arcs.

In this lesson, you will learn how to bisect a line segment, bisect an angle, and construct common angles such as  60^,  90^, and  45^ without using a protractor. You will also see why these constructions work.

These skills are important because they show how geometry can be done exactly, not by estimation. Instead of measuring, you use logical steps based on equal distances and symmetry.

1. Key Ideas You Need

A bisector cuts something into two equal parts.
A segment bisector cuts a line segment into two equal lengths.
An angle bisector divides an angle into two equal angles.
A perpendicular bisector crosses a segment at its midpoint and forms a right angle.

When doing constructions, the most important idea is this: if two points are the same distance from two endpoints of a segment, then those points help locate the midpoint. Also, if two arcs are drawn with the same compass width, they create equal distances that lead to exact constructions.

2. How to Bisect a Line Segment

Suppose you are given a segment \$\overline{AB}\$. Your goal is to find its midpoint and draw the perpendicular bisector.

Draw the segment \$\overline{AB}\$.
Open your compass to a width greater than half the length of \$AB\$.
Place the compass point on \$A\$ and draw arcs above and below the segment.
Without changing the compass width, place the compass point on \$B\$ and draw arcs that cross the first pair of arcs.
Label the intersection points of the arcs as \$P\$ and \$Q\$.
Use the straightedge to draw line \$PQ\$.

Line \$PQ\$ is the perpendicular bisector of \$\overline{AB}\$. The point where \$PQ\$ crosses \$\overline{AB}\$ is the midpoint of the segment.

Why does this work?

Because the arcs were drawn with the same compass width, point \$P\$ is the same distance from \$A\$ and \$B\$. The same is true for point \$Q\$. Any point that is equally distant from \$A\$ and \$B\$ lies on the perpendicular bisector of \$\overline{AB}\$. Since both \$P\$ and \$Q\$ have this property, the line through them must be the perpendicular bisector.

3. How to Bisect an Angle

Suppose you are given \$\angle ABC\$, where \$B\$ is the vertex. You want to split it into two equal angles.

Place the compass point on the vertex \$B\$.
Draw an arc that crosses both sides of the angle. Label the intersection points \$D\$ and \$E\$.
Without changing the compass much, place the compass point on \$D\$ and draw an arc inside the angle.
Using the same compass width, place the compass point on \$E\$ and draw another arc that crosses the first one. Label the intersection point \$F\$.
Use the straightedge to draw ray \$BF\$.

Ray \$BF\$ is the angle bisector of \$\angle ABC\$.

Why does this work?

The first arc makes \$BD = BE\$, because both are radii of the same circle centered at \$B\$. The second pair of arcs makes \$DF = EF\$. Also, \$BF\$ is shared. So triangles \$\triangle BDF\$ and \$\triangle BEF\$ are congruent by \$SSS\$. That means the two angles at \$B\$ are equal:

$$\angle DBF = \angle FBE$$

So ray \$BF\$ splits the original angle into two equal parts.

4. Constructing a \$60^\circ\$ Angle

A \$60^\circ\$ angle can be constructed using the idea of an equilateral triangle, where all sides are equal and all angles are \$60^\circ\$.

Suppose you are given a point \$A\$ and a ray \$\overrightarrow{AB}\$. You want to construct a \$60^\circ\$ angle at \$A\$.

Draw ray \$\overrightarrow{AB}\$.
Place the compass point on \$A\$ and draw an arc crossing the ray at \$B\$ or another point on the ray.
Without changing the compass width, place the compass point on that point on the ray and draw an arc that intersects the first arc. Call the new intersection point \$C\$.
Draw ray \$\overrightarrow{AC}\$.

Then \$\angle BAC = 60^\circ\$.

Why does this work?

The compass creates equal lengths, so \$AB = AC = BC\$. That makes triangle \$ABC\$ equilateral. In an equilateral triangle, all angles are equal, and the three angles must add to \$180^\circ\$:

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

5. Constructing a \$90^\circ\$ Angle

A \$90^\circ\$ angle is a right angle. One common way to construct it is by drawing a perpendicular line.

Method: Construct a perpendicular to a line at a point on the line

Suppose point \$A\$ lies on line \$\ell\$, and you want a line through \$A\$ that is perpendicular to \$\ell\$.

Place the compass point on \$A\$ and draw an arc that crosses line \$\ell\$ at two points. Call them \$B\$ and \$C\$.
Without changing the compass to a very small width, place the compass point on \$B\$ and draw an arc above or below the line.
Using the same compass width, place the compass point on \$C\$ and draw another arc that crosses the previous arc. Label the intersection \$D\$.
Draw line \$AD\$.

Line \$AD\$ is perpendicular to line \$\ell\$, so the angle formed is \$90^\circ\$.

Why does this work?

Point \$A\$ is the midpoint of \$\overline{BC}\$ because the first arc made \$AB = AC\$. Then point \$D\$ is also equally distant from \$B\$ and \$C\$, because of the equal arcs from \$B\$ and \$C\$. So line \$AD\$ is the perpendicular bisector of \$\overline{BC}\$. A perpendicular bisector forms a right angle, so the angle at \$A\$ is \$90^\circ\$.

6. Constructing a \$45^\circ\$ Angle

A \$45^\circ\$ angle can be made by first constructing a \$90^\circ\$ angle and then bisecting it.

Start with a ray or line where you want the angle.
Construct a \$90^\circ\$ angle at the chosen point.
Use the angle bisector construction on that right angle.

Since the original angle is \$90^\circ\$, the bisector divides it into two equal angles:

$$\frac{90^\circ}{2} = 45^\circ$$

7. Tips for Accurate Constructions

Do not change the compass width unless the steps tell you to.
Make arcs large enough so they clearly intersect.
Keep your pencil sharp for precise intersections.
Label important points as you go.
Do not estimate or measure with a ruler or protractor.

8. Worked Examples

Example 1: Bisecting a Segment

You are given segment \$\overline{MN}\$. Construct its perpendicular bisector.

Steps:

Draw \$\overline{MN}\$.
Set the compass to more than half of \$MN\$.
From \$M\$, draw arcs above and below the segment.
From \$N\$, with the same compass width, draw arcs crossing the first ones.
Label the arc intersections \$X\$ and \$Y\$.
Draw line \$XY\$.

Result: Line \$XY\$ crosses \$\overline{MN}\$ at its midpoint and is perpendicular to it.

Example 2: Bisecting an Angle

You are given \$\angle PQR\$. Construct its angle bisector.

Steps:

Place the compass at \$Q\$, the vertex, and draw an arc crossing both sides of the angle.
Label the crossing points \$S\$ and \$T\$.
Using the same compass width, draw arcs from \$S\$ and \$T\$ so they meet at \$U\$.
Draw ray \$QU\$.

Result: Ray \$QU\$ divides \$\angle PQR\$ into two equal angles.

Example 3: Constructing a \$60^\circ\$ Angle

You are given ray \$\overrightarrow{AB}\$. Construct a \$60^\circ\$ angle at \$A\$.

Steps:

Draw ray \$\overrightarrow{AB}\$.
With center \$A\$, draw an arc cutting the ray at \$B\$.
With center \$B\$ and the same compass width, draw an arc intersecting the first arc at \$C\$.
Draw ray \$\overrightarrow{AC}\$.

Reasoning: Since \$AB = BC = AC\$, triangle \$ABC\$ is equilateral, so \$\angle BAC = 60^\circ\$.

Example 4: Constructing a \$45^\circ\$ Angle

You need to construct a \$45^\circ\$ angle at point \$D\$.

Steps:

Draw a base ray from \$D\$.
Construct a perpendicular ray at \$D\$, making a \$90^\circ\$ angle.
Bisect that \$90^\circ\$ angle.

Reasoning: The angle bisector cuts the right angle in half:

$$90^\circ \div 2 = 45^\circ$$

9. Common Mistakes to Avoid

Using a compass width that is too small when bisecting a segment. If it is less than half the segment, the arcs will not intersect.
Changing the compass width by accident between matching arcs.
Drawing faint or incomplete arcs, which makes intersection points hard to find.
Assuming a line is bisected by sight instead of using the full construction.
Using measurement tools when the goal is exact construction by logic.

10. Quick Check for Understanding

What does a bisector do?
Why must the arcs in a perpendicular bisector construction be drawn with the same compass width?
How can you make a \$45^\circ\$ angle if you already know how to make a \$90^\circ\$ angle?
Why does constructing equal side lengths help create a \$60^\circ\$ angle?

Brief Summary

A bisector divides a segment or an angle into two equal parts. With a compass and straightedge, you can construct a segment bisector, an angle bisector, and exact angles such as \$60^\circ\$, \$90^\circ\$, and \$45^\circ\$ without measuring. These constructions work because they rely on equal distances, congruent triangles, and the properties of perpendicular bisectors.

Put what you read to the test

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

Check your understanding

Internal Division of a Line Segment in a Given Ratio

Internal Division of a Line Segment in a Given Ratio

In geometry, we sometimes need to divide a line segment into a given ratio. This means we must mark a point on the segment so that the two smaller parts are in a required proportion.

For example, if a point divides a segment in the ratio $2:3$, then one part is $2$ equal units and the other part is $3$ equal units. The whole segment is therefore split into $2+3=5$ equal parts in proportion.

This idea is called internal division because the dividing point lies between the two endpoints of the segment.

In construction work, we do not measure the segment with a ruler and calculate lengths directly. Instead, we use a compass and ruler to create equal parts and then use parallel lines to transfer those parts onto the given segment accurately.

What you will learn in this lesson:

what internal division means,
how to divide a line segment in a given ratio,
why the construction works,
and how to handle different ratios such as $1:1$, $2:3$, and $3:5$.

1. Meaning of division in a ratio

Suppose we have a line segment $AB$, and we want a point $P$ on it such that

$$\frac{AP}{PB}=m:n$$

This means

$$AP:PB = m:n$$

If $m=2$ and $n=3$, then

$$AP:PB=2:3$$

So the point $P$ should divide $AB$ into two parts where the first part is proportional to 2 and the second part is proportional to 3.

Notice that this does not always mean $AP=2\text{ cm}$ and $PB=3\text{ cm}$. It only means their lengths have the same proportion as 2 to 3.

2. Key construction idea

The construction is based on a very important geometric fact:

If a set of equal segments is marked on one ray,
and the last point is joined to the end of another segment,
then drawing a line parallel to that joining line creates proportional division on the second segment.

This is why parallel lines are used in this construction.

3. General construction: Divide $AB$ internally in the ratio $m:n$

We want to construct a point $P$ on $AB$ such that

$$AP:PB=m:n$$

Construction steps

Draw the given line segment $AB$.
From point $A$, draw a ray $AX$ making any convenient acute angle with $AB$.
On ray $AX$, mark $m+n$ equal segments using the compass. Name the points in order as $A_1, A_2, A_3, \dots\, A_{m+n}$, where all successive gaps are equal.
Join the last point $A_{m+n}$ to $B$.
Locate the point $A_m$, the $m$-th marked point from $A$.
Through $A_m$, draw a line parallel to $A_{m+n}B$. Let this parallel line meet $AB$ at point $P$.

Then point $P$ divides $AB$ internally in the ratio $m:n$.

So,

$$AP:PB=m:n$$

Why do we mark $m+n$ equal parts?

If the required ratio is $m:n$, the whole segment must be thought of as split proportionally into $m+n$ equal parts.

The point dividing the segment should be after the first $m$ parts, leaving $n$ parts for the second section.

4. Why the construction works

Let the line through $A_m$ be parallel to $A_{m+n}B$, and let it meet $AB$ at $P$.

Because the line is parallel, the triangles formed are similar in a proportional way. Equal divisions on ray $AX$ are transferred proportionally onto segment $AB$.

This gives

$$\frac{AP}{AB}=\frac{AA_m}{AA_{m+n}}=\frac{m}{m+n}$$

$$AP=\frac{m}{m+n}AB$$

Then the remaining part is

$$PB=AB-AP=AB-\frac{m}{m+n}AB=\frac{n}{m+n}AB$$

Therefore,

$$AP:PB = \frac{m}{m+n}AB : \frac{n}{m+n}AB = m:n$$

This proves that the construction is correct.

5. Important notes for construction

The ray from $A$ can be drawn at any convenient angle, but it should not be too small, or the diagram becomes crowded.
The $m+n$ parts on the ray must be equal. This is very important.
The line through $A_m$ must be parallel to $A_{m+n}B$, not just approximately slanting in the same direction.
The dividing point lies inside the segment because this is internal division.

6. Worked Example 1: Divide a segment in the ratio $1:1$

Question: Construct a point that divides line segment $AB$ internally in the ratio $1:1$.

Understanding the ratio

$$AP:PB=1:1$$

then both parts are equal. So the required point is the midpoint of $AB$.

Construction

Draw segment $AB$.
Draw a ray $AX$ from $A$.
Because $1+1=2$, mark 2 equal segments on the ray: $A_1$ and $A_2$.
Join $A_2$ to $B$.
Through $A_1$, draw a line parallel to $A_2B$ meeting $AB$ at $P$.

Result

The point $P$ divides $AB$ in the ratio $1:1$, so $P$ is the midpoint of $AB$.

Worked Example 2: Divide a segment in the ratio $2:3$

Question: Construct a point $P$ on $AB$ such that

$$AP:PB=2:3$$

Step 1: Count the total parts

Since $2+3=5$, we must mark 5 equal parts on a ray from $A$.

Construction

Draw the line segment $AB$.
From $A$, draw ray $AX$.
Mark 5 equal segments on $AX$: $A_1, A_2, A_3, A_4, A_5$.
Join $A_5$ to $B$.
Through $A_2$, draw a line parallel to $A_5B$.
Let this line meet $AB$ at $P$.

Conclusion

The point $P$ divides $AB$ internally in the ratio $2:3$.

This means the part from $A$ to $P$ corresponds to 2 equal proportional parts, and the part from $P$ to $B$ corresponds to 3 equal proportional parts.

Worked Example 3: Divide a segment in the ratio $3:5$

Question: Construct a point $P$ on segment $AB$ so that

$$AP:PB=3:5$$

Thinking first

The whole segment must be considered in $3+5=8$ equal proportional parts.

The point $P$ should come after the first 3 parts, leaving 5 parts to the other side.

Construction

Draw segment $AB$.
Draw ray $AX$ from $A$ at a convenient angle.
Mark 8 equal segments on ray $AX$: $A_1, A_2, \dots, A_8$.
Join $A_8$ to $B$.
Through $A_3$, draw a line parallel to $A_8B$.
Let it meet $AB$ at $P$.

Result

Then

$$AP:PB=3:5$$

Check using proportional idea

Since $A_3$ is the 3rd point out of 8 equal parts,

$$\frac{AP}{AB}=\frac{3}{8}$$

So the remaining part is

$$\frac{PB}{AB}=\frac{5}{8}$$

Hence,

$$AP:PB=3:5$$

Worked Example 4: Writing the method logically

In Grade 10 geometry, you may be asked not only to do the construction, but also to justify it.

Question: Describe how to divide a segment $AB$ internally in the ratio $4:1$, and explain why the method works.

Construction steps

Draw the segment $AB$.
From $A$, draw a ray $AX$.
Since $4+1=5$, mark 5 equal segments on $AX$, obtaining points $A_1, A_2, A_3, A_4, A_5$.
Join $A_5$ to $B$.
Through $A_4$, draw a line parallel to $A_5B$, meeting $AB$ at $P$.

Justification

Because $A_4P$ is parallel to $A_5B$, the divisions on the ray are transferred proportionally onto $AB$.

Since $AA_4:AA_5=4:5$, we get

$$AP:AB=4:5$$

Therefore, $AP$ is $4$ out of the total $5$ proportional parts of $AB$, leaving $1$ part for $PB$.

$$AP:PB=4:1$$

7. Common mistakes to avoid

Using only $m$ or only $n$ equal parts: always mark $m+n$ equal parts.
Choosing the wrong point on the ray: if the ratio is $m:n$, draw the parallel through the $m$-th point, not the $n$-th point from $A$.
Unequal compass steps: all marked parts on the ray must be equal.
Inaccurate parallel line: if the line is not truly parallel, the ratio will be wrong.
Reading the ratio backward: $AP:PB=2:3$ is different from $AP:PB=3:2$.

8. Quick pattern to remember

To divide $AB$ in the ratio $m:n$:

draw a ray from $A$,
mark $m+n$ equal parts on the ray,
join the last point to $B$,
through the $m$-th point draw a parallel line,
the intersection on $AB$ is the required dividing point.

9. Brief summary

Internal division of a line segment in the ratio $m:n$ means finding a point on the segment such that the two parts are proportional to $m$ and $n$.

The standard construction uses a ray from one endpoint, marks $m+n$ equal parts, joins the last point to the other endpoint, and then draws a parallel through the $m$-th point.

This method works because parallel lines create proportional segments. It allows accurate geometric construction without measuring the required lengths numerically.

Put what you read to the test

You've worked through Internal Division of a Line Segment in a Given Ratio. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Check your understanding

Constructing Similar Triangles with Scale Factors Less Than 1

Constructing Similar Triangles with Scale Factors Less Than 1

In geometry, two triangles are similar if they have the same shape but may have different sizes. Their matching angles are equal, and their matching side lengths are in the same ratio.

This lesson focuses on how to construct a smaller triangle similar to a given triangle when the scale factor is less than 1. For example, a scale factor of $\tfrac12$, $\tfrac23$, or $\tfrac34$ makes the new triangle a reduced copy of the original.

You will learn how to create this smaller triangle using geometric tools such as a ruler and compass, and by using the idea that parallel lines create proportional segments.

Big idea: If we start with a triangle and mark points on rays from one vertex in a certain ratio, then draw a line parallel to one side, the smaller triangle formed will be similar to the original triangle.

1. What does a scale factor less than 1 mean?

A scale factor tells how much a figure is enlarged or reduced. If the scale factor is less than 1, the new figure is smaller than the original.

If the scale factor is $\tfrac12$, each side becomes half as long.
If the scale factor is $\tfrac23$, each side becomes two-thirds as long.
If the scale factor is $\tfrac34$, each side becomes three-fourths as long.

If triangle $A'B'C'$ is similar to triangle $ABC$ with scale factor $k$, then

$$\frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC}=k$$

When $0<k<1$, the image triangle is a reduction.

2. Why parallel lines help us construct similar triangles

A key geometry fact is this:

If a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally. This creates a smaller triangle that is similar to the original triangle.

Suppose we have triangle $ABC$. If point $D$ is on $AB$ and point $E$ is on $AC$, and if $DE \parallel BC$, then triangle $ADE$ is similar to triangle $ABC$.

We write this as

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

and the side ratios match:

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

This is the reason the construction works. We choose a point so that one side is reduced by the correct fraction, then use a parallel line to force the rest of the triangle to shrink by the same fraction.

3. General construction method

Let the given triangle be $ABC$. We want to construct a similar triangle with scale factor $\tfrac{m}{n}$, where $m<n$.

For example, $\tfrac12$, $\tfrac23$, and $\tfrac34$ all fit this form.

Start with the given triangle $ABC$.
Choose one vertex, usually $A$, as the center of the reduction.
Draw a ray starting at $A$ in any convenient direction that does not lie on the sides of the triangle.
Mark off $n$ equal segments on this ray using a compass. Label the points $P_1, P_2, \dots, P_n$.
Connect $P_n$ to vertex $B$.
Through $P_m$, construct a line parallel to $P_nB$. Let this line meet side $AB$ at point $B'$.
Through $B'$, construct a line parallel to $BC$, or more directly, draw a line through $B'$ parallel to $BC$ so it meets $AC$ at $C'$.

Then triangle $AB'C'$ is similar to triangle $ABC$ with scale factor $\tfrac{m}{n}$.

Why does this work? Because the equal markings on the ray create the fraction $\tfrac{AP_m}{AP_n}=\tfrac{m}{n}$. The parallel line construction transfers that same fraction onto side $AB$, giving

$$\frac{AB'}{AB}=\frac{m}{n}$$

Then drawing $B'C' \parallel BC$ makes triangles $AB'C'$ and $ABC$ similar, so all corresponding sides are reduced by the same factor:

$$\frac{AB'}{AB}=\frac{AC'}{AC}=\frac{B'C'}{BC}=\frac{m}{n}$$

4. Important construction idea

In geometric construction, we do not rely on measuring side lengths with numbers. Instead, we create equal segments with a compass and use parallel lines to guarantee the correct ratio.

This is important because the goal is to justify the construction using geometry, not just arithmetic.

5. Worked Example 1: Scale factor $\tfrac12$

Problem: Construct a triangle similar to triangle $ABC$ with scale factor $\tfrac12$.

Steps:

Draw triangle $ABC$.
From vertex $A$, draw a ray $AX$.
Using the compass, mark two equal segments on the ray: $AP_1 = P_1P_2$.
Connect $P_2$ to $B$.
Through $P_1$, draw a line parallel to $P_2B$. Let it meet $AB$ at $B'$.
Through $B'$, draw a line parallel to $BC$. Let it meet $AC$ at $C'$.

Then triangle $AB'C'$ is the required reduced triangle.

Reasoning: Since $P_1$ is halfway from $A$ to $P_2$, we have

$$\frac{AP_1}{AP_2}=\frac12$$

Because the line through $P_1$ is parallel to $P_2B$, the corresponding segment on side $AB$ is also cut in the same ratio, so

$$\frac{AB'}{AB}=\frac12$$

Then $B'C' \parallel BC$, so

$$\triangle AB'C' \sim \triangle ABC$$

Therefore the whole triangle is reduced by a factor of $\tfrac12$.

6. Worked Example 2: Scale factor $\tfrac23$

Problem: Construct a triangle similar to triangle $DEF$ with scale factor $\tfrac23$.

Steps:

Draw triangle $DEF$.
From vertex $D$, draw a ray.
Mark off three equal segments on the ray: $Q_1, Q_2, Q_3$.
Connect $Q_3$ to $E$.
Through $Q_2$, draw a line parallel to $Q_3E$. Let it meet $DE$ at $E'$.
Through $E'$, draw a line parallel to $EF$. Let it meet $DF$ at $F'$.

Then triangle $DE'F'$ is similar to triangle $DEF$ with scale factor $\tfrac23$.

Why? Since the ray was divided into 3 equal parts and we used the second mark,

$$\frac{DQ_2}{DQ_3}=\frac23$$

The parallel line transfers this ratio to side $DE$, giving

$$\frac{DE'}{DE}=\frac23$$

Then the line through $E'$ parallel to $EF$ gives a smaller triangle with the same angle measures as triangle $DEF$, so the triangles are similar.

7. Worked Example 3: Explaining why the construction is valid

Problem: A student constructs triangle $AGH$ inside triangle $ABC$ so that $G$ lies on $AB$, $H$ lies on $AC$, and $GH \parallel BC$. The student claims that if $AG = \tfrac34 AB$, then triangle $AGH$ is similar to triangle $ABC$ with scale factor $\tfrac34$. Is the student correct?

Answer: Yes, the student is correct.

Since $GH \parallel BC$, corresponding angles are equal:

$\angle AGH = \angle ABC$
$\angle AHG = \angle ACB$
$\angle A$ is shared

So the triangles are similar:

$$\triangle AGH \sim \triangle ABC$$

Because $AG = \tfrac34 AB$, we have

$$\frac{AG}{AB}=\frac34$$

For similar triangles, all corresponding side lengths have the same ratio. Therefore,

$$\frac{AH}{AC}=\frac{GH}{BC}=\frac34$$

So triangle $AGH$ is a reduced triangle with scale factor $\tfrac34$.

8. Worked Example 4: Common mistake and how to fix it

Problem: A student wants a scale factor of $\tfrac25$. They draw 5 equal marks on a ray from vertex $A$, but they connect the second mark to $C$ and then draw a parallel line through the fifth mark. Will this give the correct construction?

Answer: No, this will not give the correct reduction.

To make a scale factor of $\tfrac25$, the construction must compare the 2-part point to the 5-part point in the correct order.

The usual method is:

mark 5 equal segments,
connect the 5th mark to the original side,
draw a parallel through the 2nd mark.

This creates the ratio

$$\frac25$$

If the student reverses the important points, the resulting ratio is not the intended one. In geometric constructions, the placement of the parallel line matters because it determines which fraction is created.

9. What to remember when doing the construction

Use one vertex of the triangle as the starting point for the reduction.
Write the scale factor as a fraction $\tfrac{m}{n}$ with $m<n$.
Mark $n$ equal parts on a ray.
Use the $m$-th mark to create the reduced side by drawing a line parallel to the line from the $n$-th mark.
Use a line parallel to the third side to complete the smaller triangle.
Justify your work using similarity and parallel lines, not measuring with a ruler.

10. Quick check for understanding

If the scale factor is $\tfrac35$, how many equal segments should you mark on the ray, and which mark should you use for the parallel line?

Answer: Mark 5 equal segments, and use the 3rd mark for the parallel line.

If the scale factor is $\tfrac47$, mark 7 equal segments and use the 4th mark.

11. Brief summary

To construct a triangle similar to a given triangle with a scale factor less than 1, use a ray from one vertex and divide it into equal parts. Then use parallel lines to transfer the fraction $\tfrac{m}{n}$ onto the sides of the triangle.

The key reason this works is that parallel lines create proportional segments, which guarantees similarity. This lets you build an accurate reduced triangle without depending on numerical measurement.

Put what you read to the test

You've worked through Constructing Similar Triangles with Scale Factors Less Than 1. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Check your understanding

Constructing Similar Triangles with Scale Factors Greater Than 1

Constructing Similar Triangles with Scale Factors Greater Than 1

In geometry, two figures are similar if they have the same shape but may have different sizes. For triangles, this means their matching angles are equal, and their matching side lengths are in the same ratio.

In this lesson, you will learn how to construct an enlarged triangle that is similar to a given triangle, using only geometry tools such as a ruler, compass, and straightedge. The scale factor will be greater than 1, so the new triangle will be larger than the original.

This kind of construction is important because it shows how geometry can create exact size changes without measuring lengths with numbers. Instead, we use rays, equal spacing, and parallel lines to build the correct proportions.

What does a scale factor greater than 1 mean?

A scale factor tells how much bigger or smaller a figure becomes. If the scale factor is greater than 1, the new figure is an enlargement.

For example:

A scale factor of 2 makes every side twice as long.
A scale factor of 3 makes every side three times as long.
A scale factor of $\frac{5}{2}$ makes every side $2.5$ times as long.

If triangle $A'B'C'$ is similar to triangle $ABC$ with scale factor $k>1$, then

$$\frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC}=k$$

The angles stay the same, but the side lengths are multiplied by the same number.

Main idea of the construction

To enlarge a triangle, we choose one vertex as a fixed point. Then we create a ray from that point, mark equal steps on the ray, and use parallel lines to produce a longer side in the correct ratio.

This works because parallel lines create proportional segments. That is the key geometric fact behind the construction.

Tools you need

Compass
Straightedge or ruler for drawing lines
Pencil

The basic construction method

Suppose you are given triangle $ABC$, and you want to construct a triangle similar to it with scale factor $n$, where $n$ is a whole number greater than 1.

Draw the given triangle $ABC$.
From vertex $A$, draw a new ray $AX$ going away from the triangle.
Using the compass, mark off $n$ equal segments on ray $AX$. Label the points $A_1, A_2, A_3, \dots, A_n$, so that

$$AA_1=A_1A_2=A_2A_3=\cdots=A_{n-1}A_n$$
Connect the last point $A_n$ to vertex $B$.
Through the point $A_1$, draw a line parallel to $A_nB$. Let this line meet side $AB$ or its extension in the needed relationship.

However, for enlarging the whole triangle more clearly, a more standard version is used:

Draw triangle $ABC$.
At vertex $A$, draw a ray $AX$ making any convenient angle with side $AB$.
Mark $n$ equal segments on ray $AX$: $A_1, A_2, \dots, A_n$.
Connect $A_1$ to $B$.
Through $A_n$, draw a line parallel to $A_1B$. Let it meet the extension of $AB$ at point $B'$.
Connect $A_1$ to $C$.
Through $A_n$, draw a line parallel to $A_1C$. Let it meet the extension of $AC$ at point $C'$.
The triangle $AB'C'$ is similar to triangle $ABC$ with scale factor $n$.

Because the parallel lines create matching angles, and because the equal spacing on the ray sets the ratio, we get

$$\frac{AB'}{AB}=\frac{AC'}{AC}=n$$

So triangle $AB'C'$ is an exact enlargement of triangle $ABC$.

Why this works

The construction uses the fact that if a line is drawn parallel to one side of a triangle, it creates proportional sides. This idea is sometimes called the triangle proportionality property.

Since the points on the ray are equally spaced, the distance from $A$ to $A_n$ is $n$ times the distance from $A$ to $A_1$. When the connecting lines are parallel, the triangles formed are similar.

For example, if $A_nB'$ is parallel to $A_1B$, then triangle $AA_nB'$ is similar to triangle $AA_1B$. Therefore,

$$\frac{AB'}{AB}=\frac{AA_n}{AA_1}=n$$

The same argument shows

$$\frac{AC'}{AC}=\frac{AA_n}{AA_1}=n$$

Since both enlarged sides have the same scale factor and the angle at $A$ stays the same, the new triangle is similar to the original.

Worked Example 1: Scale factor 2

Problem: Construct a triangle similar to triangle $ABC$ with scale factor 2.

Steps:

Draw triangle $ABC$.
From vertex $A$, draw a ray $AX$.
Use the compass to mark two equal segments on the ray: points $A_1$ and $A_2$.
Draw segment $A_1B$.
Through $A_2$, draw a line parallel to $A_1B$. Let it meet the extension of $AB$ at $B'$.
Draw segment $A_1C$.
Through $A_2$, draw a line parallel to $A_1C$. Let it meet the extension of $AC$ at $C'$.
Draw segment $B'C'$.

Result: Triangle $AB'C'$ is similar to triangle $ABC$, and each side is doubled.

$$\frac{AB'}{AB}=\frac{AC'}{AC}=\frac{B'C'}{BC}=2$$

Worked Example 2: Scale factor 3

Problem: Construct an enlargement of triangle $PQR$ by a scale factor of 3.

Steps:

Draw triangle $PQR$.
From vertex $P$, draw a ray $PY$.
Mark three equal segments on ray $PY$: $P_1, P_2, P_3$.
Connect $P_1$ to $Q$.
Through $P_3$, draw a line parallel to $P_1Q$. Let it meet the extension of $PQ$ at $Q'$.
Connect $P_1$ to $R$.
Through $P_3$, draw a line parallel to $P_1R$. Let it meet the extension of $PR$ at $R'$.

Why it works:

Because $PP_3=3\cdot PP_1$, the parallel-line construction makes the new sides three times the original lengths.

Therefore,

$$\frac{PQ'}{PQ}=\frac{PR'}{PR}=3$$

So triangle $PQ'R'$ is similar to triangle $PQR$.

Worked Example 3: Fractional scale factor greater than 1

Problem: Construct a triangle similar to triangle $DEF$ with scale factor $\frac{5}{2}$.

This is still an enlargement, because $\frac{5}{2}>1$.

Idea: We need a ratio of $5:2$. So instead of using 1 step and 5 steps, we compare 2 equal steps to 5 equal steps.

Steps:

Draw triangle $DEF$.
From vertex $D$, draw a ray $DZ$.
Mark five equal segments on the ray: $D_1, D_2, D_3, D_4, D_5$.
Connect $D_2$ to $E$.
Through $D_5$, draw a line parallel to $D_2E$. Let it meet the extension of $DE$ at $E'$.
Connect $D_2$ to $F$.
Through $D_5$, draw a line parallel to $D_2F$. Let it meet the extension of $DF$ at $F'$.

Why it works:

Since $DD_5:DD_2=5:2$, the similar triangles created by the parallel lines give

$$\frac{DE'}{DE}=\frac{DF'}{DF}=\frac{5}{2}$$

So triangle $DE'F'$ is similar to triangle $DEF$ with scale factor $\frac{5}{2}$.

Important construction tips

Use the same compass width each time when marking equal segments on the ray.
Make your ray long enough so all points fit clearly.
Draw parallel lines carefully. Accuracy matters in constructions.
Extend the sides of the original triangle lightly in pencil if needed.
Label points clearly so you do not confuse the original triangle with the enlarged one.

Common mistakes to avoid

Using unequal marks on the ray: If the segments are not equal, the scale factor will be wrong.
Drawing a line that is not truly parallel: Even a small error changes the size ratio.
Forgetting to extend the sides: With scale factors greater than 1, the new points usually lie beyond the original triangle.
Mixing up the comparison points: For a scale factor like $\frac{5}{2}$, be careful to connect from the 2nd mark and draw the parallel through the 5th mark.

How to justify your construction

In geometry, you should be able to explain why your construction works.

A clear justification can sound like this:

I marked equal segments on a ray from one vertex.
I connected one chosen mark to a vertex of the triangle.
I drew a parallel line through another mark.
Because the lines are parallel, the triangles formed are similar.
The ratio of the equal-step distances on the ray gives the scale factor.
Therefore, the new triangle is similar to the original triangle and is enlarged by the required factor.

Quick check questions

If the scale factor is 4, how many equal segments should you mark if you use the basic whole-number method? Answer: 4 equal segments.
If the scale factor is $\frac{3}{2}$, which two marks are important? Answer: Use the 2nd mark for the original connection and the 3rd mark for the parallel line.
If the new triangle is larger, should the new points lie inside or beyond the original sides? Answer: Usually beyond the original sides, on their extensions.

Summary

To construct a similar triangle with a scale factor greater than 1, start from one vertex and mark equal steps on a ray. Then use parallel lines to transfer the correct ratio onto the sides of the triangle.

The key idea is that parallel lines create similar triangles, and the equal step marks control the scale factor. This lets you make exact enlargements such as 2, 3, or even fractions like $\frac{5}{2}$, without using numerical measurement.

Put what you read to the test

You've worked through Constructing Similar Triangles with Scale Factors Greater Than 1. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Check your understanding

Constructing a Tangent at a Point on a Circle

Constructing a Tangent at a Point on a Circle

In geometry, a tangent to a circle is a straight line that touches the circle at exactly one point. That point is called the point of contact or point of tangency.

In this lesson, you will learn how to construct a tangent at a given point on a circle using geometry tools such as a ruler and compass. You will also learn why the method works.

This is an important construction because it connects a geometric fact you know with a practical drawing method: a tangent to a circle is perpendicular to the radius at the point of contact.

1. Key idea behind the construction

Suppose a circle has centre $O$, and $P$ is a point on the circle. If you draw the radius $OP$, then the tangent at $P$ must make a right angle with $OP$.

So the whole construction is based on this fact:

$$\text{If a line is tangent to a circle at } P, \text{ then } OP \perp \text{tangent at } P.$$

This means that to construct the tangent at $P$, we simply need to construct a line through $P$ that is perpendicular to the radius $OP$.

2. What you need

A compass
A straightedge or ruler
A pencil

You should not rely on measuring angles with a protractor or lengths with a ruler. The goal is to make the construction exactly using geometric steps.

3. The theorem you use

The main theorem is:

The tangent to a circle at any point is perpendicular to the radius drawn to that point.

In symbols, if line $l$ is tangent to the circle at $P$, and $O$ is the centre, then

$$OP \perp l$$

This theorem gives both the reason and the method for the construction.

4. Construction steps

Let the circle have centre $O$, and let $P$ be the point on the circle where you want the tangent.

Draw the radius $OP$.
At point $P$, construct a line perpendicular to $OP$.
Extend that perpendicular line on both sides of $P$.

The line you get is the tangent to the circle at $P$.

5. How to construct the perpendicular at $P$

Since $P$ lies on the line segment $OP$, you can use the standard compass construction for drawing a perpendicular to a line at a point on the line.

With centre $P$, draw a small arc cutting line $OP$ at two points. Call them $A$ and $B$.
With centres $A$ and $B$, and with the same compass width greater than half of $AB$, draw arcs above and below the line so that the arcs intersect.
Call the intersection points $X$ and $Y$.
Draw the line $XY$.

The line $XY$ is perpendicular to $AB$, and since $A$, $P$, and $B$ lie on $OP$, the line $XY$ is also perpendicular to $OP$. Because it passes through $P$, it is the tangent at $P$.

6. Why the construction works

Let us justify the method clearly.

First, $OP$ is a radius of the circle because $O$ is the centre and $P$ is on the circle.

Next, the line we construct through $P$ is perpendicular to $OP$.

By the circle theorem, the line perpendicular to the radius at the point where the radius meets the circle is a tangent.

So the constructed line touches the circle at exactly one point, $P$, and is therefore the tangent.

7. Worked Example 1: Basic construction

Question: A circle has centre $O$. Point $P$ lies on the circle. Construct the tangent at $P$.

Solution:

Join $O$ to $P$ to draw radius $OP$.
With centre $P$, draw an arc cutting $OP$ at two points $A$ and $B$.
With centres $A$ and $B$, draw arcs of equal radius above and below $OP$, meeting at $X$ and $Y$.
Draw line $XY$.

Line $XY$ passes through $P$ and is perpendicular to $OP$. Therefore, $XY$ is the tangent at $P$.

Reason: A tangent is perpendicular to the radius at the point of contact.

8. Worked Example 2: Writing a geometric justification

Question: After constructing a line through point $P$ perpendicular to radius $OP$, explain why this line is a tangent to the circle.

Solution:

Since $O$ is the centre and $P$ is on the circle, $OP$ is a radius.

The constructed line passes through $P$ and is perpendicular to $OP$.

By the theorem, the tangent to a circle at a point is perpendicular to the radius through that point.

Therefore, the constructed line is the tangent to the circle at $P$.

9. Worked Example 3: Identifying a mistake

Question: A student wants to draw a tangent at point $P$ on a circle. They draw a line through $P$ that looks as if it only touches the circle once, but they do not draw the radius $OP$ or construct a perpendicular. Why is this not a correct geometric construction?

Solution:

In a geometric construction, the result must be based on exact properties, not just appearance.

The student has not used the fact that the tangent must be perpendicular to the radius at the point of contact.

Without drawing $OP$ and constructing a perpendicular at $P$, there is no proof that the line is truly tangent.

So the method is not correct because it relies on guessing instead of using a valid construction rule.

10. Worked Example 4: Completing the reasoning

Question: Fill in the missing statement:

“To construct a tangent at point $P$ on a circle with centre $O$, draw radius $OP$. Then construct a line through $P$ that is ________ to $OP$.”

Solution:

The missing word is perpendicular.

So the completed statement is:

“To construct a tangent at point $P$ on a circle with centre $O$, draw radius $OP$. Then construct a line through $P$ that is perpendicular to $OP$.”

11. Common mistakes to avoid

Forgetting to draw the radius: You need the radius to know which line the tangent must be perpendicular to.
Guessing the tangent by eye: A construction must be exact, not estimated.
Drawing a perpendicular at the wrong point: The perpendicular must be drawn at $P$, the point on the circle.
Using measurement instead of construction: Compass-and-straightedge methods give exact geometric results.

12. Quick recall checklist

A tangent touches a circle at exactly one point.
The radius to the point of contact is perpendicular to the tangent.
To construct the tangent at $P$:
- draw radius $OP$,
- construct a perpendicular to $OP$ at $P$.

13. Brief summary

To construct a tangent at a point on a circle, first join the point to the centre to form a radius. Then draw a line through that point perpendicular to the radius. This works because the tangent to a circle is always perpendicular to the radius at the point where the tangent touches the circle.

Put what you read to the test

You've worked through Constructing a Tangent at a Point on a Circle. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Check your understanding

Constructing Tangents from an External Point

Constructing Tangents from an External Point

In geometry, a tangent to a circle is a line that touches the circle at exactly one point. That point is called the point of contact or point of tangency.

Sometimes, you are given a point outside the circle and asked to construct the tangent lines from that point to the circle using only geometric tools such as a ruler and compass. This is an important construction because it combines several ideas you already know: circles, radii, perpendicular lines, and midpoint construction.

In this lesson, you will learn how to construct tangents from an external point to a circle, and also why the construction works.

Key idea: A tangent to a circle is always perpendicular to the radius drawn to the point of tangency.

If a line touches a circle at point $T$, and $O$ is the center of the circle, then:

$$OT \perp \text{tangent at } T$$

This fact is the foundation of the construction.

The situation

Suppose:

$O$ is the center of the circle,
$P$ is a point outside the circle,
we want to draw tangent lines from $P$ to the circle.

There will usually be two tangents from the external point $P$, touching the circle at two different points, say $T_1$ and $T_2$.

Why do we use the midpoint of $OP$?

The construction works by joining the external point $P$ to the center $O$, then finding the midpoint of $OP$. Let that midpoint be $M$.

Next, we draw a new circle with center $M$ and radius $MO$ (or $MP$; these are equal because $M$ is the midpoint).

This new circle has $OP$ as a diameter, because both $O$ and $P$ lie on it and $M$ is the midpoint of $OP$.

Where this new circle cuts the original circle, we get the points of tangency.

Why is that true?

Take one intersection point and call it $T$. Since $T$ lies on the circle with diameter $OP$, angle $OTP$ is a right angle.

So:

$$\angle OTP = 90^\circ$$

That means $OT \perp PT$.

Because $OT$ is a radius of the original circle, and $PT$ is perpendicular to that radius at point $T$, the line $PT$ must be a tangent to the original circle.

This is the logical reason the construction works.

Step-by-step construction

Draw the given circle with center $O$.
Mark the external point $P$.
Join $O$ to $P$ with a straight line segment.
Construct the midpoint of $OP$. Call it $M$.
With center $M$ and radius $MO$, draw a circle.
This new circle will intersect the original circle at two points. Call them $T_1$ and $T_2$.
Join $P$ to $T_1$ and $P$ to $T_2$.
The lines $PT_1$ and $PT_2$ are the required tangents.

Tools needed

Compass
Ruler or straightedge
Pencil

Important geometric facts used

A tangent touches a circle at exactly one point.
The radius to the point of tangency is perpendicular to the tangent.
The angle in a semicircle is a right angle.
The midpoint of a line segment can be found by construction.

How to justify the construction in an exam

A good justification can be written like this:

Join the center $O$ to the external point $P$.
Let $M$ be the midpoint of $OP$.
Draw the circle with diameter $OP$ by using center $M$ and radius $MO$.
Let this circle meet the original circle at $T_1$ and $T_2$.
Since $T_1$ and $T_2$ lie on the circle with diameter $OP$, angles $OT_1P$ and $OT_2P$ are right angles.
So $OT_1 \perp PT_1$ and $OT_2 \perp PT_2$.
A line perpendicular to the radius at the point where it meets the circle is a tangent.
Therefore, $PT_1$ and $PT_2$ are tangents to the circle.

Worked Example 1: Identifying the main steps

You are given a circle with center $O$ and an external point $P$. Describe how to construct the tangent lines from $P$.

Solution

Join $O$ and $P$.
Construct the midpoint $M$ of $OP$.
Using center $M$ and radius $MO$, draw a circle.
Let the new circle cut the original circle at $T_1$ and $T_2$.
Draw $PT_1$ and $PT_2$.

These are the required tangent lines.

Worked Example 2: Explaining why the construction works

After constructing points $T_1$ and $T_2$, explain why $PT_1$ is a tangent.

Solution

Point $T_1$ lies on the circle with diameter $OP$. Therefore, angle $OT_1P$ is a right angle.

So:

$$\angle OT_1P = 90^\circ$$

This means $OT_1 \perp PT_1$.

Since $OT_1$ is a radius of the original circle, and a tangent is perpendicular to the radius at the point of contact, $PT_1$ is a tangent to the circle.

The same argument shows that $PT_2$ is also a tangent.

Worked Example 3: Completing a reasoning question

A student says, “I joined the external point $P$ to the center $O$, found the midpoint $M$, and drew a circle centered at $M$ through $O$ and $P$. It cut the original circle at $A$ and $B$. So $PA$ and $PB$ are tangents.”

Complete the student’s reasoning.

Solution

The circle centered at $M$ has $OP$ as a diameter.
Since $A$ and $B$ lie on this circle, angles $OAP$ and $OBP$ are right angles.
Therefore, $OA \perp PA$ and $OB \perp PB$.
Because $OA$ and $OB$ are radii of the original circle, $PA$ and $PB$ are tangents to the original circle.

Worked Example 4: Common mistake check

A student joins $O$ to $P$, but instead of finding the midpoint of $OP$, they draw a circle centered at $O$ passing through $P$. Will this help construct the tangents?

Solution

No. That circle is not the special circle needed for the construction.

The correct extra circle must have diameter $OP$. To make that happen, its center must be the midpoint of $OP$, not $O$.

The reason the construction works is that points on the circle with diameter $OP$ make a right angle with $O$ and $P$. If the circle is centered at $O$ instead, that right-angle property is not guaranteed.

Common mistakes to avoid

Not using the midpoint of $OP$: The second circle must be based on the midpoint so that $OP$ becomes a diameter.
Forgetting to join $O$ and $P$: You need this segment before you can find its midpoint.
Joining the tangency points to the center instead of the external point: The required tangent lines must start from the external point $P$.
Assuming one tangent only: From a point outside a circle, there are usually two tangents.
Using measurement instead of construction: In geometric constructions, rely on ruler and compass steps, not numerical measuring.

Quick recall: How to construct the midpoint of $OP$

Place the compass at $O$ and draw arcs above and below the segment $OP$.
Without changing the compass width, place it at $P$ and draw arcs that cross the first pair.
Join the two arc-intersection points.
This line is the perpendicular bisector of $OP$, and where it meets $OP$ is the midpoint $M$.

What you should remember

To construct tangents from an external point $P$, first join $P$ to the center $O$.
Find the midpoint of $OP$.
Draw the circle with diameter $OP$.
Its intersections with the original circle are the points of tangency.
Joining these points to $P$ gives the tangent lines.

Brief summary

Constructing tangents from an external point depends on one key fact: the tangent is perpendicular to the radius at the point of contact. By drawing a circle with diameter $OP$, where $O$ is the center and $P$ is the external point, we create right angles at the intersection points. Those right angles show exactly where the tangent lines must touch the circle.

Put what you read to the test

You've worked through Constructing Tangents from an External Point. Try answering a few questions to see what stuck — and what might deserve a quick reread before you move on.

Check your understanding

Geometric Proofs of Construction Validity

Geometric Proofs of Construction Validity means showing, with logic, that a compass-and-straightedge construction really creates the figure it is supposed to create.

In geometry, it is not enough to say, “It looks right,” or “I measured it.” A valid construction must be justified by geometric facts such as congruent circles, equal radii, perpendicular bisectors, and angle bisectors.

This lesson explains how to prove that a construction works. You will learn how to connect each compass step to a geometric reason, and then use those reasons to write a clear proof.

Why proofs matter in constructions

When you do a construction, your tools are limited. A straightedge draws a line through points. A compass draws a circle with a chosen center and radius. Because these tools are exact in geometry, every mark you make has a meaning.

For example, if you draw a circle centered at point $A$ through point $B$, then every point on that circle is exactly the same distance from $A$ as point $B$ is. That means every radius of that circle has length $AB$.

A proof of construction validity explains why those exact relationships force the final figure to have the required property.

The basic idea of a construction proof

Most construction proofs follow this pattern:

State what was constructed.
List the key facts created by the compass or straightedge steps.
Use known geometry theorems or definitions.
Conclude that the construction has the desired result.

In other words, you must answer this question: Why do these steps guarantee the result?

Important geometric facts often used in construction proofs

All radii of the same circle are congruent.
If a point lies on a circle centered at $A$, then its distance from $A$ equals the radius.
The set of points equidistant from the endpoints of a segment lies on the perpendicular bisector of that segment.
The set of points equidistant from the sides of an angle lies on the angle bisector of that angle.
Vertical angles are congruent.
If two sides and the included angle of one triangle are congruent to those of another triangle, then the triangles are congruent (SAS).
If corresponding parts of congruent triangles match, then their angles or sides are equal (CPCTC).

How compass steps create exact facts

Each time you draw an arc or circle, you create equal lengths. This is one of the most important ideas in construction proofs.

Suppose you draw a circle centered at $A$ with radius $AB$. If point $P$ lies on that circle, then

$$AP = AB$$

because both are radii of the same circle.

If you also draw a circle centered at $B$ with radius $AB$, and point $P$ is where the circles intersect, then

$$AP = AB \quad \text{and} \quad BP = AB$$

$$AP = BP$$

This tells us that $P$ is equidistant from $A$ and $B$. That fact is often the key to proving a perpendicular bisector or an equilateral triangle.

Writing a construction proof clearly

A good proof should refer to the construction steps directly. For example:

“Point $P$ is an intersection point of circles centered at $A$ and $B$.”
“Since $P$ lies on the circle centered at $A$, $AP = AB$.”
“Since $P$ lies on the circle centered at $B$, $BP = AB$.”
“Therefore $AP = BP$.”

Then connect that fact to a theorem:

“Because $P$ is equidistant from $A$ and $B$, point $P$ lies on the perpendicular bisector of $\overline{AB}$.”

Example 1: Proving an equilateral triangle construction

Construction: Given segment $\overline{AB}$, construct an equilateral triangle on $\overline{AB}$.

Steps:

Draw a circle centered at $A$ with radius $AB$.
Draw a circle centered at $B$ with radius $AB$.
Let $C$ be an intersection point of the two circles.
Draw segments $\overline{AC}$ and $\overline{BC}$.

Goal: Prove that triangle $ABC$ is equilateral.

Proof:

Because $C$ lies on the circle centered at $A$, $AC = AB$.

Because $C$ lies on the circle centered at $B$, $BC = AB$.

$$AC = AB \quad \text{and} \quad BC = AB$$

Therefore, by the transitive property,

$$AC = BC = AB$$

All three sides of triangle $ABC$ are congruent, so triangle $ABC$ is equilateral.

Why this proof works: The construction did not measure side lengths. Instead, the circles guaranteed equal lengths because radii in the same circle are congruent.

Example 2: Proving a perpendicular bisector construction

Construction: Given segment $\overline{AB}$, construct its perpendicular bisector.

Steps:

Set the compass to a width greater than half of $AB$.
Draw arcs centered at $A$ above and below the segment.
Without changing the compass width, draw arcs centered at $B$ that intersect the first arcs at points $P$ and $Q$.
Draw line $PQ$.

Goal: Prove that line $PQ$ is the perpendicular bisector of $\overline{AB}$.

Proof:

Since $P$ is on an arc centered at $A$ and on an arc centered at $B$, we have

$$PA = PB$$

Similarly, since $Q$ is on both arcs,

$$QA = QB$$

So both $P$ and $Q$ are equidistant from $A$ and $B$.

Any point equidistant from the endpoints of a segment lies on the perpendicular bisector of that segment. Therefore, both $P$ and $Q$ lie on the perpendicular bisector of $\overline{AB}$.

There is exactly one line through points $P$ and $Q$, so line $PQ$ is that perpendicular bisector.

Thus, line $PQ$ is perpendicular to $\overline{AB}$ and passes through its midpoint.

Why this proof works: The proof depends on the idea of a locus: the set of all points equidistant from $A$ and $B$ forms the perpendicular bisector of $\overline{AB}$.

Example 3: Proving an angle bisector construction

Construction: Given $\angle ABC$, construct its angle bisector.

Steps:

Draw an arc centered at $B$ that intersects rays $BA$ and $BC$ at points $D$ and $E$.
Using the same compass width, draw arcs centered at $D$ and $E$ that intersect at point $F$ inside the angle.
Draw ray $BF$.

Goal: Prove that ray $BF$ bisects $\angle ABC$.

Proof:

Because $D$ and $E$ lie on the same arc centered at $B$,

$$BD = BE$$

Because $F$ lies on arcs centered at $D$ and $E$ drawn with the same compass width,

$$DF = EF$$

Also, $BF$ is a common side, so

$$BF = BF$$

Now compare triangles $\triangle DBF$ and $\triangle EBF$:

$$BD = BE, \quad DF = EF, \quad BF = BF$$

Therefore, the triangles are congruent by SSS.

So their corresponding angles at $B$ are congruent:

$$\angle DBF = \angle FBE$$

Point $D$ lies on ray $BA$, and point $E$ lies on ray $BC$. Therefore,

$$\angle ABF = \angle FBC$$

So ray $BF$ divides $\angle ABC$ into two congruent angles. Therefore, ray $BF$ is the angle bisector of $\angle ABC$.

Why this proof works: The construction creates two triangles with matching side lengths. Triangle congruence then shows the two angles are equal.

Example 4: Proving the construction of a line perpendicular to a given line through a point on the line

Construction: Given line $\ell$ and point $P$ on $\ell$, construct a line through $P$ perpendicular to $\ell$.

One common method:

With center $P$, draw an arc that intersects line $\ell$ at points $A$ and $B$.
Using the same compass width, draw arcs centered at $A$ and $B$ above the line, intersecting at point $C$.
Draw line $PC$.

Goal: Prove that $PC \perp \ell$.

Proof:

Since the first arc was centered at $P$, points $A$ and $B$ lie on that circle. Therefore,

$$PA = PB$$

So $P$ is equidistant from $A$ and $B$.

Since point $C$ lies on equal-radius arcs centered at $A$ and $B$,

$$CA = CB$$

So $C$ is also equidistant from $A$ and $B$.

Any point equidistant from $A$ and $B$ lies on the perpendicular bisector of $\overline{AB}$. Therefore, both $P$ and $C$ lie on the perpendicular bisector of $\overline{AB}$.

The line through $P$ and $C$ must be that perpendicular bisector. Thus, line $PC$ is perpendicular to $\overline{AB}$.

Because $A$ and $B$ lie on line $\ell$, segment $\overline{AB}$ is part of line $\ell$. Therefore,

$$PC \perp \ell$$

A strategy for proving any construction

When you are given a construction and asked to justify it, use these questions:

What points were created by circle intersections?
What equal lengths come from radii or equal compass widths?
Do those equal lengths show a point is equidistant from two endpoints or two sides?
Can I use triangle congruence?
What final definition do I need? For example, perpendicular bisector, angle bisector, equilateral triangle, midpoint, or perpendicular line.

Common mistakes to avoid

Do not say “it looks equal.” Visual appearance is not proof.
Do not rely on ruler measurements. Construction proofs must use geometric properties, not approximate measurements.
Do not skip the reason a length is equal. If you claim $AC = AB$, explain that both are radii of the same circle.
Do not jump to the conclusion too early. First show the needed relationships, then state the theorem or definition.

Helpful proof language

Here are sentence starters you can use:

“Because point $P$ lies on the circle centered at ...”
“Since the compass width was unchanged ...”
“Therefore the two lengths are congruent.”
“This means the point is equidistant from ...”
“By the definition of perpendicular bisector ...”
“By triangle congruence ...”
“Therefore the construction is valid.”

Quick check for understanding

If two arcs are drawn with the same compass width from endpoints $A$ and $B$, and they intersect at point $P$, why is $P$ important?

The answer is that $P$ is equidistant from $A$ and $B$, so $P$ lies on the perpendicular bisector of $\overline{AB}$. That single idea appears in many geometric construction proofs.

Summary

A geometric proof of construction validity shows that a compass-and-straightedge construction must work because of geometry facts, not because of how it looks.

The most common proof tools are:

equal radii from circles,
points equidistant from segment endpoints,
triangle congruence, and
definitions of bisectors, perpendicular lines, and equilateral triangles.

When writing a proof, connect each construction step to a fact, then use a theorem or definition to reach the conclusion. If you do that carefully, you can prove that your construction is mathematically exact.

Put what you read to the test

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

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