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 𝐴𝐸 from vertex 𝐴 so that it is parallel to 𝐶𝐷. Then, we extend 𝐵𝐶 beyond point 𝐶 until 𝐵𝐶 intersects with 𝐴𝐸 at point 𝐸.

Since 𝐶𝐷 is the bisector of angle 𝐵𝐶𝐷, we know that the two marked angles in the diagram at vertex 𝐶 are congruent. Also, since 𝐴𝐸𝐶𝐷,

  • 𝐵𝐶𝐷 and 𝐵𝐸𝐴 are corresponding angles, so they are congruent,
  • 𝐷𝐶𝐴 and 𝐶𝐴𝐸 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 𝐵𝐴𝐸, we have two parallel lines, 𝐴𝐸 and 𝐷𝐶 that are cut by two transversals, 𝐵𝐴 and 𝐵𝐸, so we have 𝐵𝐷𝐷𝐴=𝐵𝐶𝐶𝐸.

Finally, we notice that 𝐴𝐶𝐸 is an isosceles triangle because 𝑚𝐶𝐴𝐸=𝑚𝐶𝐸𝐴, which means 𝐶𝐸=𝐶𝐴. Hence, 𝐵𝐷𝐷𝐴=𝐵𝐶𝐶𝐴.

Rewriting 𝐵𝐷=𝐷𝐵 and 𝐵𝐶=𝐶𝐵 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, 𝐴𝐷 bisects 𝐵𝐴𝐶, 𝐵𝐷=8, 𝐷𝐶=11, and the perimeter of 𝐴𝐵𝐶 is 57. Determine the lengths of 𝐴𝐵 and 𝐴𝐶.

Answer

We are given that 𝐴𝐷 is the bisector of an interior angle at 𝐴 in triangle 𝐴𝐵𝐶. 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 𝐷𝐶𝐵𝐷=𝐴𝐶𝐴𝐵.

Since we are given 𝐵𝐷=8 and 𝐷𝐶=11, this means 118=𝐴𝐶𝐴𝐵.

We are also given that the perimeter of triangle 𝐴𝐵𝐶 is 57. Recall that the perimeter of a polygon is the sum of the lengths of the sides of the polygon. This leads to 𝐴𝐵+𝐵𝐶+𝐴𝐶=57.

Since point 𝐷 divides 𝐵𝐶 into two smaller line segments, 𝐵𝐷 and 𝐷𝐶, we can substitute 𝐵𝐶=𝐵𝐷+𝐷𝐶 to write 𝐴𝐵+𝐵𝐷+𝐷𝐶+𝐴𝐶=57.

Substituting 𝐵𝐷=8 and 𝐷𝐶=11, we have

𝐴𝐵+8+11+𝐴𝐶=57𝐴𝐵+𝐴𝐶=38.(1)

We recall the equation obtained previously, 118=𝐴𝐶𝐴𝐵, which can be rearranged to

𝐴𝐶=118𝐴𝐵.(2)

Substituting this expression into equation (1) and simplifying, we have 𝐴𝐵+𝐴𝐶=38𝐴𝐵+118𝐴𝐵=38198𝐴𝐵=38𝐴𝐵=819×38=16.

We can substitute this value into equation (2) to obtain 𝐴𝐶=118𝐴𝐵=118×16=22.

Hence, 𝐴𝐵=16,𝐴𝐶=22.

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 𝐴𝐶=10, 𝐶𝐷=6, 𝐴𝐵=𝑥+9, and 𝐵𝐷=𝑥+5, find the numerical value of 𝑥.

Answer

We are given that 𝐴𝐷 is the bisector of the interior angle at 𝐴 in triangle 𝐴𝐵𝐶. 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 𝐷𝐶𝐵𝐷=𝐴𝐶𝐴𝐵.

We can substitute given lengths and expressions into this equation to write 6𝑥+5=10𝑥+9.

Cross multiplying and simplifying, we obtain 6(𝑥+9)=10(𝑥+5)6𝑥+54=10𝑥+506𝑥10𝑥=50544𝑥=4𝑥=1.

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 𝐴𝐵𝐶, where we are given a point 𝐷 on side 𝐵𝐶.

If we are given that the lengths of the line segments satisfy 𝐷𝐵𝐷𝐶=𝐴𝐵𝐴𝐶, we know that point 𝐷 lies on the bisector of the interior angle at vertex 𝐴 in triangle 𝐴𝐵𝐶.

We now turn our attention to an analogous bisector theorem for an exterior angle of a triangle.

Theorem: Exterior Angle Bisector Theorem

Consider 𝐵𝐶𝐸, which is an exterior angle of triangle 𝐴𝐵𝐶 at vertex 𝐶, is bisected by 𝐶𝐷 that intersects the extension of the side of the triangle opposite 𝐶, which is 𝐴𝐵, at a point 𝐷, as shown in the following diagram.

We have the following identity: 𝐷𝐵𝐷𝐴=𝐶𝐵𝐶𝐴.

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 𝐷𝐸 from vertex 𝐷 so that it is parallel to 𝐵𝐶. This parallel line intersects with the extension of side 𝐴𝐶 at a point 𝐸, as shown in the diagram below.

We note that angles 𝐶𝐷𝐸 and 𝐷𝐶𝐵 are alternate angles with respect to the two parallel lines; hence, they are congruent. This makes triangle 𝐸𝐷𝐶 isosceles. Since 𝐵𝐶 and 𝐷𝐸 are parallel, we can see that triangles 𝐴𝐵𝐶 and 𝐴𝐷𝐸 are similar. In particular, this is because the two triangles share an angle at 𝐴, and the other two pairs of angles are corresponding angles with respect to the parallel lines.

This means that for some positive constant 𝑟, we have 𝐴𝐵𝐴𝐷=𝑟,𝐴𝐶𝐴𝐸=𝑟,𝐵𝐶𝐷𝐸=𝑟.

If we take the equation 𝐴𝐶𝐴𝐸=𝑟, we can write 𝐴𝐶=𝑟𝐴𝐸=𝑟(𝐴𝐶+𝐶𝐸).

Since we know that 𝐸𝐷𝐶 is isosceles, we have 𝐶𝐸=𝐷𝐸. This leads to

𝐴𝐶=𝑟(𝐴𝐶+𝐷𝐸).(3)

Now, we can rearrange the equation 𝐵𝐶𝐷𝐸=𝑟 to write 𝐷𝐸=𝐵𝐶𝑟. Substituting this into the right-hand side of equation (3), we obtain 𝑟(𝐴𝐶+𝐷𝐸)=𝑟𝐴𝐶+𝐵𝐶𝑟=𝑟𝐴𝐶+𝐵𝐶.

Hence,

𝐴𝐶=𝑟𝐴𝐶+𝐵𝐶.(4)

Now, we can substitute 𝑟=𝐴𝐵𝐴𝐷 into (4) to write 𝐴𝐶=𝐴𝐵𝐴𝐷×𝐴𝐶+𝐵𝐶.

Multiplying both sides of the equation by 𝐴𝐷 and rearranging, we obtain 𝐴𝐶𝐴𝐷=𝐴𝐵𝐴𝐶+𝐵𝐶𝐴𝐷𝐴𝐶𝐴𝐷𝐴𝐵𝐴𝐶=𝐵𝐶𝐴𝐷𝐴𝐶(𝐴𝐷𝐴𝐵)=𝐵𝐶𝐴𝐷𝐴𝐶𝐵𝐷=𝐵𝐶𝐴𝐷.

This is equivalent to the desired ratio 𝐷𝐵𝐷𝐴=𝐶𝐵𝐶𝐴, 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 𝐴𝐵=60, 𝐴𝐶=40, and 𝐵𝐶=31, what is 𝐶𝐷?

Answer

In the diagram, it is indicated that 𝐴𝐷 is the bisector of the exterior angle at 𝐴 of triangle 𝐴𝐵𝐶. We recall the exterior angle bisector theorem that gives the identity 𝐷𝐶𝐷𝐵=𝐴𝐶𝐴𝐵.

Since 𝐷𝐵=𝐷𝐶+𝐶𝐵, we can write 𝐷𝐶𝐷𝐶+𝐶𝐵=𝐴𝐶𝐴𝐵.

We are given the lengths 𝐴𝐵=60, 𝐴𝐶=40, and 𝐵𝐶=31, which we can substitute into the equation above to write 𝐷𝐶𝐷𝐶+31=4060.

The equation above has one unknown length, 𝐷𝐶, which is the same as length 𝐶𝐷 that we are looking for. Cross multiplying and simplifying the resulting equation, we obtain 60𝐶𝐷=40(𝐶𝐷+31)60𝐶𝐷=40𝐶𝐷+124020𝐶𝐷=1240𝐶𝐷=124020=62.

Hence, 𝐶𝐷=62.

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 𝐴𝐵=30cm, 𝐵𝐶=40cm, and 𝐴𝐶=45cm, find the ratio between the areas of 𝐴𝐸𝐷 and 𝐴𝐸𝐶.

Answer

Let us consider the areas of the two triangles 𝐴𝐸𝐷 and 𝐴𝐸𝐶. We know that the area of a triangle is given by 0.5××.baselengthcorrespondingheight

The height of both triangles is given by the length of a line segment from vertex 𝐴 that intersects perpendicularly with line 𝐸𝐶. Let us call this height , as given in the following diagram.

The base length of triangle 𝐴𝐸𝐷 is 𝐸𝐷, and the base length of triangle 𝐴𝐸𝐶 is 𝐸𝐶. Hence, areaofareaof𝐴𝐸𝐷=12𝐸𝐷×,𝐴𝐸𝐶=12𝐸𝐶×.

Since 12 is a common factor in both areas, the ratio of the two areas can be written as 𝐸𝐷𝐸𝐶.

Neither of these lengths is provided, so let us return to our given diagram. We can see that 𝐴𝐷 is the bisector of the interior angle at 𝐴 of triangle 𝐴𝐵𝐶 and also that 𝐴𝐸 is the bisector of the exterior angle at 𝐴 of triangle 𝐴𝐵𝐶. 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: 𝐷𝐶𝐷𝐵=𝐴𝐶𝐴𝐵,𝐸𝐵𝐸𝐶=𝐴𝐵𝐴𝐶.

We can substitute the given lengths 𝐴𝐵=30cm, 𝐵𝐶=40cm, and 𝐴𝐶=45cm to write

𝐷𝐶𝐷𝐵=4530=32,𝐸𝐵𝐸𝐶=3045=23.(5)(6)

Since we know that 𝐸𝐶=𝐸𝐵+𝐵𝐶, we can substitute this expression into equation (6) above to write 𝐸𝐵𝐸𝐵+𝐵𝐶=23.

We can cross multiply this equation and substitute 𝐵𝐶=40cm to obtain 3𝐸𝐵=2(𝐸𝐵+𝐵𝐶)3𝐸𝐵=2𝐸𝐵+2𝐵𝐶3𝐸𝐵2𝐸𝐵=2×40𝐸𝐵=80.cm

We also know that 𝐵𝐷+𝐷𝐶=𝐵𝐶, which means 𝐷𝐶=𝐵𝐶𝐵𝐷=40𝐷𝐵. Substituting this expression into equation (5), 40𝐷𝐵𝐷𝐵=32.

Simplifying gives 2(40𝐷𝐵)=3𝐷𝐵802𝐷𝐵=3𝐷𝐵80=2𝐷𝐵+3𝐷𝐵5𝐷𝐵=80𝐷𝐵=16.cm

Finally, 𝐷𝐶=40𝐷𝐵=4016=24.cm

Now, we can compute 𝐸𝐷=𝐸𝐵+𝐵𝐷=80+16=96,𝐸𝐶=𝐸𝐷+𝐷𝐶=96+24=120.cmcm

This means the ratio 𝐸𝐷𝐸𝐶 can be written as 96120. Dividing each part by 24, this ratio is equivalent to 45.

Hence, the ratio between the areas of 𝐴𝐸𝐷 and 𝐴𝐸𝐶 is 45.

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 𝐴𝐵𝐶, if 𝐶𝐷 is the angle bisector of angle 𝐶, then we have

Let us prove this theorem. We begin by adding a circle circumscribing the triangle above and also adding intersecting point 𝐸 between line 𝐶𝐷 and the circle.

In the diagram above, we know that the two green angles at vertex 𝐶 are congruent since 𝐶𝐷 is the bisector of 𝐴𝐶𝐵. We also note that the angles at vertices 𝐵 and 𝐸 are inscribed angles subtended by common arc 𝐴𝐶. 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, 𝐶𝐵𝐷 and 𝐶𝐸𝐴 share two pairs of congruent angles, which means that these are similar triangles. Hence, 𝐶𝐷𝐶𝐴=𝐶𝐵𝐶𝐸.

Writing 𝐶𝐸=𝐶𝐷+𝐷𝐸, we have 𝐶𝐷𝐶𝐴=𝐶𝐵𝐶𝐷+𝐷𝐸.

Cross multiplying and simplifying as before, we obtain

𝐶𝐷(𝐶𝐷+𝐷𝐸)=𝐶𝐵𝐶𝐴𝐶𝐷+𝐶𝐷𝐷𝐸=𝐶𝐵𝐶𝐴𝐶𝐷=𝐶𝐵𝐶𝐴𝐶𝐷𝐷𝐸.(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 𝐶𝐷𝐷𝐸 on the right-hand side of the equation above should be replaced with 𝐷𝐵𝐷𝐴. In order to justify this substitution, we need to observe another pair of similar triangles.

In the diagram above, we can see that 𝐵𝐶𝐸 and 𝐵𝐴𝐸 are inscribed angles subtended by the same arc, which means that they are congruent. Then, 𝐶𝐵𝐷 and 𝐴𝐸𝐷 share two pairs of congruent angles, which means that they are similar. This gives us 𝐶𝐷𝐷𝐴=𝐷𝐵𝐷𝐸, which leads to 𝐶𝐷𝐷𝐸=𝐷𝐵𝐷𝐴.

We can now substitute the expression of the right-hand side of equation (7) to write 𝐶𝐷=𝐶𝐵𝐶𝐴𝐷𝐵𝐷𝐴.

Taking the positive square root of both sides of this equation gives 𝐶𝐷=𝐶𝐵𝐶𝐴𝐷𝐵𝐷𝐴, 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 𝐴𝐵𝐶, 𝐴𝐵=76cm, 𝐴𝐶=57cm, and 𝐵𝐷=52cm. Given that 𝐴𝐷 bisects 𝐴 and intersects 𝐵𝐶 at 𝐷, determine the length of 𝐴𝐷.

Answer

We recall that if 𝐴𝐷 bisects 𝐴 in triangle 𝐴𝐵𝐶, we have 𝐴𝐷=𝐴𝐶𝐴𝐵𝐷𝐶𝐷𝐵.

From the question, we know the lengths of 𝐴𝐶, 𝐴𝐵, and 𝐷𝐵. Therefore, to find 𝐴𝐷, we must first find 𝐷𝐶. 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 𝐷𝐶𝐵𝐷=𝐴𝐶𝐴𝐵.

Substituting the given lengths into this equation, 𝐷𝐶52=5776𝐷𝐶=577652=39.cm

Now, we can substitute this length and the other given lengths into the formula for the length of the angle bisector: 𝐴𝐷=𝐴𝐶𝐴𝐵𝐷𝐶𝐷𝐵=57763952=48.cm

Hence, the length of 𝐴𝐷 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 𝐴𝐵𝐶, where we are given a point 𝐷 on side 𝐵𝐶.
    If we are given that the lengths of the line segments satisfy 𝐷𝐵𝐷𝐶=𝐴𝐵𝐴𝐶, we know that point 𝐷 lies on the bisector of the interior angle at vertex 𝐴 in triangle 𝐴𝐵𝐶.
  • Consider 𝐵𝐶𝐸, which is an exterior angle of triangle 𝐴𝐵𝐶 at vertex 𝐶, is bisected by 𝐶𝐷 that intersects the extension of the side of the triangle opposite 𝐶, which is 𝐴𝐵, at a point 𝐷, as shown in the following diagram.
    We have the following identity: 𝐷𝐵𝐷𝐴=𝐶𝐵𝐶𝐴.
  • In any triangle 𝐴𝐵𝐶, if 𝐶𝐷 is the angle bisector of angle 𝐶, then we have

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