Geometry (A): READ: Angle Bisector Reading

Angle Bisectors in Triangles

Learning Objectives

Construct the bisector of an angle.
Apply the Angle Bisector Theorem to identify the point of concurrency of the perpendicular bisectors of the sides (the incenter).
Use the Angle Bisector Theorem to solve problems involving the incenter of triangles.

Introduction

In our last lesson we examined perpendicular bisectors of the sides of triangles. We found that we were able to use perpendicular bisectors to circumscribe triangles. In this lesson we will learn how to inscribe circles in triangles. In order to do this, we need to consider the angle bisectors of the triangle. The bisector of an angle is the ray that divides the angle into two congruent angles.

Here is an example of an angle bisector in an equilateral triangle.

Angle Bisector Theorem and its Converse

We can prove the following pair of theorems about angle bisectors.

Angle Bisector Theorem: If a point is on the bisector of an angle, then the point is equidistant from the sides of the angle.

Before we proceed with the proof, let’s recall the definition of the distance from a point to a line. The distance from a point to a line is the length of the line segment that passes through the point and is perpendicular to the original line.

Proof. Consider $\angle{ROS}$ with angle bisector $\overline{OP}$ , and segments $\overline{PX}$ and $\overline{PY}$ , perpendicular to each side through point $P$ as follows:

We will show that $\overline{PX} \cong \overline{PY}$ .

Since $\overline{OP}$ is the bisector of $\angle{ROS}$ , then $\angle{XOP} \cong \angle{YOP}$ by the definition of angle bisector. In addition, since $\overline{PX}$ and $\overline{PY}$ are perpendicular to the sides of $\angle{ROS}$ , then $\angle{PXO}$ and $\angle{PYO}$ are right angles and thus congruent. Finally, $\overline{OP} \cong \overline{OP}$ by the reflexive property.
By the AAS postulate, we have $\triangle PXO \cong \triangle PYO$ .
So $\overline{PX} \cong \overline{PY}$ by CPCTC (corresponding parts of congruent triangles are congruent). $\blacklozenge$

Therefore $P$ is equidistant from each side of the angle. And since $P$ represents any point on the angle bisector, we can say that every point on the angle bisector is equidistant from the sides of the angle.

We can also prove the converse of this theorem.

Converse of the Angle Bisector Theorem: If a point is in the interior of an angle and equidistant from the sides, then it lies on the bisector of the angle.

Proof. Consider $\angle{ROS}$ with points $X$ and $Y$ and segment $\overline{OP}$ such that $\overline{PX} \cong \overline{PY}$ as follows:

As the distance to each side is given by the lengths of $\overline{PX}$ and $\overline{PY}$ respectively, we have that $\overline{PX}$ and $\overline{PY}$ are perpendicular to sides $\overline{RO}$ and $\overline{SO}$ respectively.
Note that $\overline{PO}$ is the hypotenuse of right triangles $\triangle XOP$ and $\triangle YOP$ Hence, since $\overline{PX} \cong \overline{PY}$ , and $\overline{PO}\cong\overline{PO}$ , and $\angle{PXO}$ and $\angle{PYO}$ are right angles, then the triangles are congruent by Theorem 4-6.
$\angle{POX} \cong \angle{POY}$ by CPCTC.
Hence, point $P$ lies on the angle bisector of $\angle{ROS}$ . $\blacklozenge$

Notice that we just proved the Angle Bisector Theorem (If a point is on the angle bisector then it is equidistant from the sides of the angle) and we also proved the converse of the Angle Bisector theorem (If a point is equidistant from the sides of an angle then it is on the angle bisector of the triangle). When we have proven both a theorem and its converse we say that we have proven a biconditional statement. We can put the two conditional statements together using if and only if: "A point is on the angle bisector of an angle if and only if it is equidistant from the sides of the triangle."

Angle Bisectors in a Triangle

Concurrency of Angle Bisectors Theorem: The angle bisectors of a triangle intersect in a point that is equidistant from the three sides of the triangle called the incenter.

Example 1

Inscribe the following triangle using a compass and a straightedge.

1. Draw triangle $\triangle ABC$ with your straightedge.

2. Use your compass to construct the angle bisectors and find the point of concurrency $X$ .

3. Use your compass to construct the circle that inscribes $\triangle ABC$ .

Example 2

Inscribe a circle within the following triangle using The Geometer’s Sketchpad.

We can use the commands of GSP to construct the incenter and corresponding circle as follows:

1. Open a new sketch and construct triangle $\triangle ABC$ using the Segment Tool.

2. You can construct the angle bisectors of the angles by first designating the angle by selecting the appropriate vertices (e.g., to select the angle at vertex $A$ , select points $B$ , $A$ and $C$ in order) and then choosing Construct Angle Bisector from the Construct menu. After bisecting two angles, construct the point of intersection by selecting each angle bisector and choosing Intersection from the Construct menu. (Recall from our proof of the concurrency of angle bisectors theorem that we only need to bisect two of the angles to find the incenter.)

3. You are now ready to construct the circle. Recall that the radius of the circle must be the distance from $X$ to each side – our figure above does not include that segment. However, we do not need to construct the perpendicular line segments as we did to prove Theorem 5-7. Sketchpad will measure the distance for us.

4. To measure the distance from $X$ to each side, select point $X$ and one side of the triangle. Choose Distance from the Measure menu. This will give you the radius of your circle.

5. We are now ready to construct the circle. Select point $X$ and the distance from $X$ to the side of the triangle. Select “Construct circle by center + radius” from the Construct menu. This will give the inscribed circle within the triangle.

Lesson Summary

In this lesson we:

Defined the angle bisector of an angle.
Stated and proved the Angle Bisector Theorem.
Solved problems using the Angle Bisector Theorem.
Constructed angle bisectors and the inscribed circle with compass and straightedge, and with Geometer’s Sketchpad.

Points to Consider

How are circles related to triangles, and how are triangles related to circles? If we draw a circle first, what are the possibilities for the triangles we can circumscribe? In later chapters we will more carefully define and work with the properties of circles.

The following questions are for your own review. The answers are listed below for you to check your work and understanding.

Review Questions

Construct the incenter of $\triangle ABC$ and the inscribed circle for each of the following triangles using a straightedge, compass, and Geometer's Sketchpad.

In the last lesson we found that we could circumscribe some kinds of quadrilaterals as long as opposite angles were supplementary. Use Geometer's Sketchpad to explore the following quadrilaterals and see if you can inscribe them by the angle bisector method.
1. a square
2. a rectangle
3. a parallelogram
4. a rhombus
5. From your work in a-d, what condition must hold in order to circumscribe a quadrilateral?
Consider equilateral triangle $\triangle ABC$ . Construct the angle bisectors of the triangle and the incenter $X$ . Connect the incenter to each vertex so that the line segment intersects the side opposite the angle as follows.

As with circumcenters, we get six congruent $30^\circ-60^\circ-90^\circ$ triangles. Now connect the points that intersect the sides. What kind of figure do you get?
True or false: An incenter can also be a circumcenter. Illustrate your reasoning with a drawing.
Consider the situation described in exercise 4 for the case of an isosceles triangle. What can you conclude about the six triangles that are formed?
Consider line segment $\overline{AB}$ with coordinates $A(4,4)$ , $B(8,4)$ . Suppose that we wish to find points $C$ and $D$ so that the resulting quadrilateral can be either circumscribed or inscribed. What are some possibilities for locating points $C$ and $D?$
Using a piece of tracing or Patty Paper, construct an equilateral triangle. Bisect one angle by folding one side onto another. Unfold the paper. What can you conclude about the fold line?
Repeat exercise 8 with an isosceles triangle. What can you conclude about repeating the folds?
What are some other kinds of polygons where you could use Patty Paper to bisect an angle into congruent figures?
Given:
$\overline{ST}$ is the perpendicular bisector of $\overline{QR}$ .

$\overline{QT}$ is the perpendicular bisector of $\overline{SP}$ .

Prove:

$\overline{PQ} \cong \overline{RS}$ .
Given:
$\overline{PQ}$ bisects $\angle{XQR}$ .

$\overline{PR}$ bisects $\angle{QRZ}$ .

Prove:

$\overline{PY}$ bisects $\angle{XYZ}$ .

Review Answers

1. Yes
2. No: Bisectors are not concurrent at a point.
3. No: Bisectors are not concurrent at a point.
4. Yes
5. Angle bisectors must be concurrent.
Equilateral triangle
The statement is true in the case of an equilateral triangle. In addition, for squares the statement is also true.
We do not get six congruent triangles as before. But we get four congruent triangles and a separate pair of congruent triangles. In addition, if we connect the points where the bisectors intersect the sides, we get an isosceles triangle.
From our previous exercises we saw that we could inscribe and circumscribe some but not all types of quadrilaterals. Drawing from those exercises, we see that we could circumscribe and inscribe a square. So, locating the points at $C(4, 8)$ and $D(8, 8)$ is one such possibility. Similarly, we could locate the points at $C(6, 6)$ and $D(6, 2)$ and get a kite that can be inscribed but not circumscribed.
The fold line divides the triangle into two congruent triangles and thus is a line of symmetry for the triangle. Note that the same property will hold by folding at each of the remaining angles.
There is only one fold line that divides the triangle into congruent triangles, the line that folds the angle formed by the congruent sides.
Any regular polygon will have this property. For example, a regular pentagon:

Last modified: Monday, June 28, 2010, 3:53 PM