In this explainer, we will learn how to use

the angle bisector theorem and its converse to find a missing side

length in a triangle.

In a triangle, a bisector of an interior angle intersects the side

opposite the angle, which splits the side into two line segments of smaller

lengths. What can we say about the ratio of the lengths of these

line segments? This question is answered by the angle bisector theorem,

which we now state.

Theorem: Interior Angle Bisector Theorem

If an interior angle of a triangle is bisected, the bisector divides the

opposite side into segments whose lengths have the same ratio as the

lengths of the noncommon adjacent sides of the respective bisected angle.

That is,

Let us prove this theorem. We begin by drawing

from vertex so that it is

parallel to . Then, we extend

beyond point until

intersects with

at point .

Since is the bisector of angle

, we know that the two marked angles in the

diagram at vertex are congruent. Also, since

,

and are

corresponding angles, so they are congruent,

and are alternate angles,

so they are congruent.

This leads to the fact that all four marked angles in the diagram above are

congruent. We note that in , we have

two parallel lines, and

that are cut by two transversals,

and

, so we have

Finally, we notice that is an

isosceles triangle because , which means . Hence,

Rewriting and leads

to the statement of the theorem.

In our first example, we will apply the interior angle bisector

theorem to find missing lengths in a triangle.

Example 1: Finding Unknown Side Lengths in a Triangle Using the Angle Bisector

Theorem

In the figure, bisects

, ,

,

and the perimeter of is 57. Determine the lengths of and

.

Answer

We are given that is the

bisector of an interior angle at in

triangle . We recall the interior angle bisector

theorem. If an interior angle of a triangle is bisected, the bisector

divides the opposite side into segments whose lengths have the same

ratio as the lengths of the noncommon adjacent sides of the respective

bisected angle. In the given diagram, this means

Since we are given and , this means

We are also given that the perimeter of triangle

is 57. Recall that the perimeter of a polygon is the sum of the lengths

of the sides of the polygon. This leads to

Since point divides

into two smaller line segments,

and

, we can substitute

to write

Substituting and , we have

We recall the equation obtained previously,

which can be rearranged to

Substituting this expression into equation (1)

and simplifying, we have

We can substitute this value into equation (2)

to obtain

Hence,

Let us consider another example where we will identify an unknown term

by using this theorem.

Example 2: Using the Angle Bisector Theorem to Form and Solve

an Algebraic Equation

Given that , ,

, and , find

the numerical value of .

Answer

We are given that is the

bisector of the interior angle at in triangle

. We recall the interior angle bisector theorem

that relates the ratio of the lengths of line segments related to an

angle bisector. In this diagram, this theorem tells us

We can substitute given lengths and expressions into this equation to

write

Cross multiplying and simplifying, we obtain

In the previous two examples, we applied the theorem for the ratio of line

segments related to the bisector of an interior angle of a triangle. The

converse of this theorem is also true. More precisely, consider a triangle

, where we are given a point on

side .

If we are given that the lengths of the line segments satisfy

we know that point lies on the bisector of the interior angle

at vertex in triangle .

We now turn our attention to an analogous bisector theorem for an exterior

angle of a triangle.

Theorem: Exterior Angle Bisector Theorem

Consider , which is an exterior angle of triangle

at vertex , is bisected by

that intersects the extension of the side of

the triangle opposite , which is

, at a point , as shown in the

following diagram.

We have the following identity:

The proof of this theorem is more complex than that of the interior bisector

theorem, although the ideas are similar. Let us prove this theorem. We

begin by drawing from vertex

so that it is parallel to

. This parallel line intersects with the

extension of side at a point

, as shown in the diagram below.

We note that angles and

are alternate angles with respect to the

two parallel lines; hence, they are congruent. This makes triangle

isosceles. Since

and

are parallel, we can see that

triangles and are similar. In

particular, this is because the two triangles share an angle at

, and the other two pairs of angles are corresponding angles

with respect to the parallel lines.

This means that for some positive constant , we have

If we take the equation , we can write

Since we know that is isosceles,

we have . This leads to

Now, we can rearrange the equation

to write . Substituting this into

the right-hand side of equation (3), we obtain

Hence,

Now, we can substitute into

(4) to write

Multiplying both sides of the equation by and rearranging, we obtain

This is equivalent to the desired ratio

which proves the theorem.

Let us consider an example where we will apply the exterior angle bisector

theorem to find a missing length in a triangle.

Example 3: Finding Unknown Side Lengths in a Triangle Using the Angle Bisector

Theorem

Given that , ,

and , what is ?

Answer

In the diagram, it is indicated that

is the bisector of the exterior

angle at of triangle . We

recall the exterior angle bisector theorem that gives the identity

Since , we can write

We are given the lengths , ,

and , which we can substitute into the equation above

to write

The equation above has one unknown length, ,

which is the same as length that we are

looking for. Cross multiplying and simplifying the resulting

equation, we obtain

Hence, .

In the next example, we will apply both theorems for the interior and

exterior angle bisectors for a triangle to find the ratio of the areas of

two triangles.

Example 4: Finding the Ratio between the Areas of Two Triangles Using the Angle

Bisector Theorem

If ,

, and

, find the ratio between the

areas of and

.

Answer

Let us consider the areas of the two triangles

and . We know that the area of a triangle is

given by

The height of both triangles is given by the length of a line segment

from vertex that intersects perpendicularly with

line . Let us call this height

, as given in the following diagram.

The base length of triangle is

, and the base length of triangle

is . Hence,

Since is a common factor in both areas,

the ratio of the two areas can be written as

Neither of these lengths is provided, so let us return to our given

diagram. We can see that is

the bisector of the interior angle at of triangle

and also that

is the bisector of the exterior angle at of triangle

. We recall the theorems regarding the ratio of line

segments related to the interior and exterior angle bisectors of a triangle. The theorems tell us the following identities:

We can substitute the given lengths ,

, and

to write

Since we know that , we can

substitute this expression into equation (6) above to write

We can cross multiply this equation and substitute

to obtain

We also know that , which means

. Substituting this

expression into equation (5),

Simplifying gives

Finally,

Now, we can compute

This means the ratio can be written as

. Dividing each part by 24, this ratio is

equivalent to .

Hence, the ratio between the areas of

and

is .

So far, we have discussed the ratio of lengths of

line segments related to the bisector of an interior or exterior angle of

a triangle. We will now consider a theorem that deals with the length of

the angle bisector line segment.

Theorem: Length of the Angle Bisector in a Triangle

In any triangle , if is the

angle bisector of angle , then we have

Let us prove this theorem. We begin by adding a circle circumscribing the

triangle above and also adding intersecting point

between line and the circle.

In the diagram above, we know that the two green angles at vertex

are congruent since

is the bisector of

. We also note that the angles at vertices

and are inscribed angles subtended

by common arc . We recall that all angles subtended by the common

arc have equal measures, which tells us that the two red angles in the

diagram are congruent. Then, and

share two pairs of congruent angles, which means that

these are similar triangles. Hence,

Writing , we have

Cross multiplying and simplifying as before, we obtain

We note here that taking the positive square root of both sides of this

equation will not produce the desired formula. In particular, the term

on the right-hand side of the equation

above should be replaced with . In order to justify this

substitution, we need to observe another pair of similar triangles.

In the diagram above, we can see that and

are inscribed angles subtended by the

same arc, which means that they are congruent. Then,

and

share two pairs of congruent

angles, which means that they are similar. This gives us

which leads to

We can now substitute the expression of the right-hand side of equation

(7) to write

Taking the positive square root of both sides of this equation gives

which proves the theorem.

In the final example, we will apply this theorem to find the length of the

bisector of an interior angle of a triangle.

Example 5: Finding an Unknown Side Length in a Triangle Using the Angle Bisector

Theorem

In the triangle ,

,

,

and . Given that bisects

and intersects

at , determine the length of

.

Answer

We recall that if bisects

in triangle ,

we have

From the question, we know the lengths of

,

, and

. Therefore, to find

, we must

first find . To find this length, we recall another theorem regarding

the angle bisector: if an interior angle of a triangle is bisected, the

bisector divides the opposite side into segments whose lengths have the same

ratio as the lengths of the noncommon adjacent sides of the respective

bisected angle. In the given diagram, this means

Substituting the given lengths into this equation,

Now, we can substitute this length and the other given lengths into

the formula for the length of the angle bisector:

Hence, the length of is

48 cm.

Let us consider an extension to this angle bisector length theorem, where we consider the length of the bisector

of an exterior angle of a triangle.

Theorem: Length of the Exterior Angle Bisector in a Triangle

In any triangle , if is the angle bisector of angle

, then we have

This theorem can be proved by using the previous theorem in combination with the other angle bisector theorems we

have learned so far, although we will not include the proof in this explainer.

Let us consider an example where we must use the exterior angle bisector theorems to find missing lengths in a

triangle.

Example 6: Finding an Unknown Side Length in a Triangle Using the Exterior Angle Bisector Theorem

Given that is a triangle in which , find the value of each of

and .

Answer

Recall that if bisects the exterior angle at

in triangle , then we have

We note that in the diagram is equal to . We have been given that

and , but the lengths and

need to be calculated by finding .

To find , we can use the exterior angle bisector theorem, which gives us another relation

between the lengths of sides in a triangle where the exterior angle is bisected. Namely, they are related by the

following ratio:

Substituting the given lengths in, we have

We can solve this for by cross multiplying and rearranging to get

Now that we have , let us return to the original equation. Substituting in the given lengths

and , we have

In conclusion, and .

Let us finish by recapping a few important concepts from this explainer.

Key Points

If an interior angle of a triangle is bisected, the bisector divides

the opposite side into segments whose lengths have the same ratio as the

lengths of the noncommon adjacent sides of the respective bisected angle.

That is,

The converse of the theorem above is also true. More precisely, consider a

triangle , where we are given a point

on side .

If we are given that the lengths of the line segments satisfy

we know that point lies on the bisector of the

interior angle at vertex in triangle

.

Consider , which is an exterior angle of triangle

at vertex , is bisected by

that intersects the extension of the side of the

triangle opposite , which is

,

at a point , as shown in the following diagram.

We have the following identity:

In any triangle , if

is the angle bisector of angle , then we have

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