# Lesson Explainer: Angle Bisector Theorem and Its Converse Mathematics

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

 𝐴𝐵+8+11+𝐴𝐶=57𝐴𝐵+𝐴𝐶=38. (1)

We recall the equation obtained previously, which can be rearranged to

 𝐴𝐶=118𝐴𝐵. (2)

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

 𝐴𝐶=𝑟(𝐴𝐶+𝐷𝐸). (3)

Now, we can rearrange the equation to write . Substituting this into the right-hand side of equation (3), we obtain

Hence,

 𝐴𝐶=𝑟𝐴𝐶+𝐵𝐶. (4)

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

 𝐷𝐶𝐷𝐵=4530=32,𝐸𝐵𝐸𝐶=3045=23. (5)(6)

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

 𝐶𝐷(𝐶𝐷+𝐷𝐸)=𝐶𝐵⋅𝐶𝐴𝐶𝐷+𝐶𝐷⋅𝐷𝐸=𝐶𝐵⋅𝐶𝐴𝐶𝐷=𝐶𝐵⋅𝐶𝐴−𝐶𝐷⋅𝐷𝐸. (7)

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 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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