When a perpendicular bisector cuts a line segment at a right angle and into two congruent segments, we can use the perpendicular bisector theorem and the converse of the perpendicular bisector theorem to solve problems in angles and triangles.

Perpendicular bisector theorem

If a point is on the perpendicular bisector of a line segment, then it is equidistant from the end points of the line segment.

Segment M N with a point P on M N. Another segment O P is drawn perpendicular to M N. Points Q, R, S and T are on O P. For each point, dashed segment are drawn from the point to M and N.

Converse of perpendicular bisector theorem

If a point is equidistant from the end points of a line segment, then it is on the perpendicular bisector of that line segment

Equidistant

The same distance from two or more objects

To construct the perpendicular bisector of a segment, we will:

Identify the segment we want to bisect.
Open the compass width to just past half the segment's length and draw an arc from one endpoint that extends to both sides of the segment.
Without changing the compass width, draw another arc from the other endpoint that intersects the original arc on both sides of the segment.
Label the intersection of the arcs with points.
Connect the points.

A diagram showing the 5 steps of constructing the bisector of a segment. Speak to your teacher for more information.

Examples

Example 1

Recall that the Pythagorean theorem states that given a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of its legs lengths.

Prove the perpendicular bisector theorem.

Worked Solution

Create a strategy

To prove this theorem, construct a segment \overline{PM} \perp \overline{AB} such that M is the midpoint of \overline{AB}.

Since both \triangle AMP and \triangle BMP are right angled triangles with AM=BM (given) and \overline{PM} common to both we can use the Pythagorean theorem and substitution to find the relationship between PA and PB.

Triangle A P B with point M on side A B. A segment is drawn from P to M perpendicular to A B. Segments A M and M B are congruent.

Apply the idea

To prove: PA=PB
	Statements	Reasons
1.	AM=BM	Given
2.	\triangle AMP is a right triangle	\overline{PM}\perp\overline{AB}
3.	\triangle BMP is a right triangle	\overline{PM}\perp\overline{AB}
4.	PA^{2}=AM^{2}+PM^{2}	Pythagorean theorem in \triangle AMP
5.	PB^{2}=BM^{2}+PM^{2}	Pythagorean theorem in \triangle BMP
6.	PB^{2}=AM^{2}+PM^{2}	Substitution property of equality
7.	PA^{2}=PB^{2}	Transitive property of equality
8.	PA=PB	Square root property

Example 2

Construct perpendicular bisector \overleftrightarrow{CD} through \overline{AB}.

Worked Solution

Create a strategy

One approach to constructing the perpendicular bisector through \overline{AB} is using a compass and straightedge.

Apply the idea

1. Construct an arc centered at point A so that the arc is longer than half of \overline{AB}.

Segment A B. An arc centered at A is drawn.

2.Construct an arc centered at point B using the same compass setting, such that this arc intersects the arc constructed in Step 1 at two distinct points. Label the two points C and D.

Segment A B. Arcs centered at A and B are drawn.

3. Draw \overleftrightarrow{CD}. Label the point M where \overleftrightarrow{CD} intersects \overline{AB}. M is the midpoint of \overline{AB}.

Segment A B. Arcs centered at A and B are drawn. A line is drawn using the points of intersection of the two arcs. The point of intersection of A B and the line is labeled M.

\overleftrightarrow{CD}\perp \overline{AB} at the midpoint of \overline{AB}, so \overleftrightarrow{CD} is the perpendicular bisector of \overline{AB}.

Example 3

Find the value of x.

A triangle with leg lengths of 15 and 2 x plus 3. A segment is drawn from the apex of the triangle to a point on the base of the triangle, and that is perpendicular to the base. This segment divides the base of the triangle into 2 congruent segments.

Worked Solution

Create a strategy

We can see from the diagram that the vertical line is the perpendicular bisector because it cuts the horizontal line in half and is perpendicular to it.

Apply the idea

Since we have a perpendicular bisector, we can apply the perpendicular bisector theorem which tells us that the lengths with measures of 15 and 2x+3 are equal.

\displaystyle 15	\displaystyle =	\displaystyle 2x+3	Perpendicular bisector theorem
\displaystyle 12	\displaystyle =	\displaystyle 2x	Subtract 3 from both sides
\displaystyle 6	\displaystyle =	\displaystyle x	Divide both sides by 2

Idea summary

A segment, line, or ray that is perpendicular to a line segment at its midpoint is a perpendicular bisector.

Angle bisectors

Recall that an angle bisector is a line, segment or ray that divides an angle into two congruent angles. When an angle bisector cuts an angle into two congruent angles, we can use the angle bisector theorem and the converse of the angle bisector theorem to solve problems in angles and triangles.

Angle bisector theorem

If a point is on an angle bisector, then it is equidistant from the two sides of the angle

An angle and a segment. The first endpoint of the segment is on the vertex of the angle, and the second endpoint is in the interior of the angle. The segment divides the angle into two equal angles. From the second vertex of the segment, one segment is drawn to and perpendicular to one of the legs of the angle, and another segment is drawn to and perpendicular to the other leg of the angle.

Converse of angle bisector theorem

If a point is in the interior of an angle and is equidistant from the sides of the angle, then it lies on the angle bisector

Recall, to construct the bisector of an angle, we:

Identify the angle we want to bisect.
With the compass point on the vertex of the angle, use the compass to draw an arc that intersects both legs.
Label the intersections with points.
With the compass point on one of the points from Step 3, draw an arc that passes halfway through the interior of the angle.
With the compass point on the other point from Step 3, draw an arc that passes halfway through the interior of the angle and intersects the first arc.
Label the intersection of the arcs drawn in parts 4 and 5 with a point.
Draw a line that connects the vertex of the angle and the point added in Step 6.

A diagram showing the 7 steps of constructing the bisector of an angle. Speak to your teacher for more information.

Examples

Example 4

Find x. Justify your answer.

Worked Solution

Create a strategy

The vertical line segment is an angle bisector by the converse of angle bisector theorem which states that if a point is in the interior of an angle and is equidistant from the sides of the angle, then it lies on the angle bisector. Use the theorem to determine x.

Apply the idea

Since the vertical line segment is an angle bisector, x=23.

Example 5

P is the incenter of the triangle.

Triangle A B C with incenter P. Angle B has a measure of 76 degrees and angle C has a measure of 62 degrees.

Determine m \angle BAP.

Worked Solution

Create a strategy

Since P is the incenter of the triangle, it is the point of concurrency of the angle bisectors. So we know that \overline{AP} bisects \angle BAC, and therefore m\angle BAP = \dfrac{1}{2} m\angle BAC.

We can use the measures of the two known angles to find the measure of \angle BAC.

Apply the idea

\displaystyle m\angle BAC + m\angle ABC + m\angle ACB	\displaystyle =	\displaystyle 180\degree	Triangle interior angle sum
\displaystyle m\angle BAC + 76\degree + 62\degree	\displaystyle =	\displaystyle 180\degree	Substitute m\angle{ABC}=76\degree and m\angle{ACB}=62\degree
\displaystyle m\angle BAC +138\degree	\displaystyle =	\displaystyle 180\degree	Combine like terms
\displaystyle m\angle BAC	\displaystyle =	\displaystyle 42\degree	Subtract 138\degree from both sides

Then we have

\displaystyle m\angle BAP	\displaystyle =	\displaystyle \frac{1}{2} m\angle BAC	\overline{AP} bisects \angle BAC
\displaystyle m\angle BAP	\displaystyle =	\displaystyle \frac{1}{2} (42\degree)	Substitute m\angle{BAC}=42\degree
\displaystyle m\angle BAP	\displaystyle =	\displaystyle 21 \degree	Evaluate the multiplication

Determine m \angle BPC.

Worked Solution

Create a strategy

Again, since P is the incenter of the triangle, it is the point of concurrency of the angle bisectors. We can use this to find the measures of \angle PBC and \angle PCB. and then use those to determine the measure of \angle BPC.

Apply the idea

\displaystyle m\angle PBC	\displaystyle =	\displaystyle \frac{1}{2} m\angle ABC	Definition of angle bisector
	\displaystyle =	\displaystyle \frac{1}{2} \left(76 \degree\right)	Substitute m\angle{ABC}=76\degree
	\displaystyle =	\displaystyle 38 \degree	Evaluate the multiplication

and

\displaystyle m\angle PCB	\displaystyle =	\displaystyle \frac{1}{2} m\angle ACB	Definition of angle bisector
	\displaystyle =	\displaystyle \frac{1}{2} \left(62 \degree\right)	Substitute angle measure
	\displaystyle =	\displaystyle 31 \degree	Simplify expression

Then we have

\displaystyle m\angle BPC + m\angle PBC + m\angle PCB	\displaystyle =	\displaystyle 180\degree	Triangle interior angle sum
\displaystyle m\angle BPC + 38\degree + 31\degree	\displaystyle =	\displaystyle 180\degree	Substitute m\angle{BPC}=38\degree and m\angle{PCB}=31\degree
\displaystyle m\angle BPC + 69\degree	\displaystyle =	\displaystyle 180\degree	Combine like terms
\displaystyle m\angle BPC	\displaystyle =	\displaystyle 111\degree	Subtract 69\degree from both sides

Example 6

Draw a triangle and construct the inscribed circle.

Worked Solution

Create a strategy

The point of concurrency of the angle bisectors of a triangle is called the incenter because it is the center of the inscribed circle of the triangle.

To find the incenter, we need to draw at least two of the three angle bisectors. Two is sufficient as the three lines are concurrent, but three would be more precise if constructing by hand.

Once we have the incenter, we need to find the radius of the circle which will be the shortest (perpendicular) distance to the triangle, so we can draw a perpendicular line from the incenter to one of the sides and use that as the compass width for the inscribed circle.

Apply the idea

Draw a triangle and construct one of the angle bisectors.
Construct at least one more angle bisector.
Find the point of intersection of the two angle bisectors, this is the incenter. Then construct a line that is perpendicular to one of the sides of the triangle that goes through the incenter.
Set the compass width to the distance between the incenter and where the perpendicular line crossed the triangle side. Draw a circle with center at the incenter.
Erase all construction marks.

Example 7

A landscaper used a coordinate plane to plan the layout of a garden. The plan includes the three paths shown below, that pass through the points A \left(30,0\right), B \left(30,45\right) and C \left(5,26.25\right). The landscaper wants to place a fountain at an equal distance from each path.

-5

x \left(\text{ft}\right)

-5

y \left(\text{ft}\right)

How could they find the coordinates of the fountain?

Worked Solution

Create a strategy

We need to identify whether perpendicular bisectors or angle bisectors will help us solve the problem. The goal in this problem is to find a location equal distance from each path.

Apply the idea

The landscaper can find the location of the fountain by considering the triangle formed by the paths and finding the intersection of the angle bisectors of the vertices of that triangle.

What is the location of the fountain?

Worked Solution

Create a strategy

Copy the diagram and use constructions to find the location of the fountain. We only need two angle bisectors to find the incenter of the paths.

Apply the idea

-5

x \left(\text{ft}\right)

-5

y \left(\text{ft}\right)

The fountain is located at X\left(20,25\right) on the coordinate plane.

Idea summary

The incenter is the point of concurrency of the angle bisectors of a triangle. It is called the incenter because it is the center of the inscribed circle of the triangle.

The incenter will be equidistant from each side of a triangle.

To construct the inscribed circle of a triangle, we can construct angle bisectors on two angles and a perpendicular bisector. Then, we can draw a circle using the incenter and the point where the perpendicular line crosses the triangle's side as the radius.

4.04 Angle and perpendicular bisectors

Ideas

Perpendicular bisectors

Exploration

Examples

Example 1

Example 2

Example 3

Angle bisectors

Examples

Example 4

Example 5

Example 6

Example 7

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