# Lesson Explainer: Angle Bisector Theorem and Its Converse

SSC – Maths2 E – Chap1 – V7 – Converse of Angle Bisector Theorem – Similarity
SSC – Maths2 E – Chap1 – V7 – Converse of Angle Bisector Theorem – Similarity

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,

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
.

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 .

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 ?

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
.

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

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
.

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 .

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