### Video Transcript

In this video, weβll learn how to
use the angle bisector theorem and its converse to find a missing side length in a
triangle. A bisector of an interior angle of
a triangle intersects the opposing side to the angle. The opposing side is split into two
line segments. And a useful theorem concerning the
ratio of the lengths of these line segments is the interior angle bisector
theorem. Weβll consider this with some
examples together with the exterior angle bisector theorem. And weβll look at how to calculate
the length of the angle bisector in a triangle.

The interior angle bisector theorem
says that if an interior angle of a triangle is bisected, that is, the angle split
into two smaller angles of equal measure, then the bisector divides the opposite
side into segments whose lengths have the same ratio as the lengths of the noncommon
adjacent sides of the bisected angle. What this means with respect to our
triangle is that the length π·π΅ divided by the length π·π΄ is the same as the
length πΆπ΅ over the length πΆπ΄.

To prove this, letβs make some room
and redraw our triangle. We begin by adding a new line from
the vertex π΄ parallel to the line πΆπ·. We then extend the line segment
π΅πΆ until it meets this new line. And the two lines intersect at the
point πΈ. Since the line πΆπ· is the bisector
of the angle at the point πΆ, we know that the two angles in the diagram at vertex
πΆ are congruent. That is, theyβre the same. Now, since line segments π΄πΈ and
πΆπ· are parallel, then angles π΅πΆπ· and π΅πΈπ΄ are congruent. And where angles π΅πΆπ· and π΅πΈπ΄
are corresponding angles, angles π·πΆπ΄ and πΆπ΄πΈ are alternate angles and
therefore also congruent. And this means that all four marked
angles in the diagram are congruent.

And since triangles π΅πΆπ· and
π΅πΈπ΄ share two congruent angles, theyβre similar triangles, then it follows that
the ratio π΅π· over π΅π΄ is the same as the ratio π΅πΆ over π΅πΈ. And we could rewrite this as shown
since π΅π΄ is the same as π΅π· plus π·π΄ and π΅πΈ is the same as π΅πΆ plus πΆπΈ. If we then cross multiply and
distribute the parentheses, we have π΅π· multiplied by π΅πΆ plus π΅π· multiplied by
πΆπΈ is equal to π΅πΆ multiplied by π΅π· plus π΅πΆ multiplied by π·π΄. Subtracting π΅π· multiplied by π΅πΆ
from both sides, this then reduces to π΅π· multiplied by πΆπΈ is π΅πΆ multiplied by
π·π΄. And now dividing both sides by π·π΄
multiplied by πΆπΈ, we can divide the numerator and denominator on the left by πΆπΈ
and the numerator and denominator on the right by π·π΄. And making some room, we have π΅π·
divided by π·π΄ is equal to π΅πΆ divided by πΆπΈ.

Finally, we notice that the
triangle π΄πΆπΈ is an isosceles triangle. And this means that πΆπΈ is equal
to πΆπ΄. And we can replace this in our
equation. So we have π΅π· over π·π΄ is equal
to π΅πΆ over πΆπ΄. And simply rewriting π΅π· as π·π΅
and π΅πΆ as πΆπ΅, we have a formula π·π΅ over π·π΄ is equal to πΆπ΅ over πΆπ΄. Letβs look at an example of how we
use this theorem to find missing lengths in a triangle.

In the figure, the line segment
π΄π· bisects the angle π΅π΄πΆ, π΅π· is equal to eight, π·πΆ is equal to 11, and the
perimeter of the triangle π΄π΅πΆ is 57. Determine the lengths of the line
segments π΄π΅ and π΄πΆ.

Weβre told that the line segment
π΄π· bisects the angle at π΄. And to answer this question, weβre
going to use the interior angle bisector theorem. This says that if the interior
angle of a triangle is bisected, the bisector divides the opposite sides into
segments. And the lengths of these segments
have the same ratio as the lengths of the noncommon sides adjacent to the bisected
angle. What this means in our case is that
the length π·πΆ divided by the length π΅π· is the same as the length π΄πΆ divided by
the length π΄π΅.

Weβre given that π΅π· is equal to
eight and π·πΆ is equal to 11. And this means that 11 over eight
is equal to π΄πΆ over π΄π΅. Weβre also told that the perimeter
of the triangle is 57. And since the perimeter is the sum
of the lengths of the sides, we have π΄π΅ plus π΅πΆ plus π΄πΆ is 57. And since we know that π΅πΆ is πΆπ·
plus π·π΅, and thatβs 11 plus eight, we can replace π΅πΆ with 19. Now, if we subtract 19 from both
sides, we have π΄π΅ plus π΄πΆ is equal to 38. Now if we recall our equation from
earlier, thatβs 11 over eight is equal to π΄πΆ over π΄π΅, we can find π΄πΆ in terms
of π΄π΅ by multiplying both sides by π΄π΅. This gives us 11 over eight π΄π΅ is
equal to π΄πΆ.

And now replacing this in our
second equation, we have π΄π΅ plus 11 over eight π΄π΅ is equal to 38. Now collecting our terms on the
left, we have 19 over eight π΄π΅ is equal to 38. Now, multiplying both sides by
eight and dividing by 19, we have π΄π΅ is 38 multiplied by eight over 19 so that
π΄π΅ is equal to 16. We can now use this in our equation
for π΄πΆ so that we have 11 over eight multiplied by 16 is equal to π΄πΆ. 11 over eight multiplied by 16 is
22. Therefore, π΄πΆ is 22. Hence, the length of the line
segment π΄πΆ is 22 and that of π΄π΅ is 16.

In this example, 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.

Consider the triangle π΄π΅πΆ where
weβre given a point π· on the side π΅πΆ. If weβre given also that the
lengths of the line segments satisfy the equation π·π΅ over π·πΆ is equal to π΄π΅
over π΄πΆ, then we know that the point π· lies on the bisector of the interior angle
π΅π΄πΆ. Now, letβs look at an analogous
bisector theorem for an exterior angle of a triangle.

Consider the bisector of an
exterior angle at vertex πΆ of a triangle π΄π΅πΆ intersecting an extension of the
side of the triangle opposite to the angle at a point π·. We then have the identity π·π΅ over
π·π΄ is equal to πΆπ΅ over πΆπ΄. That is, the ratio of the length
π·π΅ to the length π·π΄ is equal to the ratio of the length πΆπ΅ to the length
πΆπ΄. And this is the exterior angle
bisector theorem. Although we wonβt prove this here,
we know that the ideas are similar, though more complex, to those used in proving
the interior angle bisector theorem. Letβs consider an example of how to
apply the exterior angle bisector theorem to find a missing length in a
triangle.

Given that π΄π΅ is equal to 60,
π΄πΆ is 40, and π΅πΆ is 31, what is πΆπ·?

The diagram indicates that the two
angles shown are congruent. That is, theyβre the same number of
degrees or radians. The line segment π΄π· is then the
bisector of the exterior angle at π΄ of the triangle π΄π΅πΆ. We recall that the exterior angle
bisector theorem gives us the identity π·πΆ over π·π΅ is equal to π΄πΆ over π΄π΅,
that is, that the ratio of the length π·πΆ to the length π·π΅ is the same as the
ratio of the length π΄πΆ to the length π΄π΅. Now, we know that the length π·π΅
is equal to π·πΆ plus πΆπ΅. So we can replace this in our
formula. And we have π·πΆ over π·πΆ plus
πΆπ΅ is equal to π΄πΆ over π΄π΅.

Weβre given the length π΄π΅ is
equal to 60, π΅πΆ is equal to 31, and π΄πΆ is equal to 40. Into our equation then, the
equation has only one unknown; thatβs π·πΆ. And our equation is then π·πΆ
divided by π·πΆ plus 31 is 40 over 60. And notice since the length π·πΆ is
the same as πΆπ·, this is what weβre looking for. So now letβs solve our equation for
π·πΆ. Multiplying both sides by 60 and
also π·πΆ plus 31, we have 60π·πΆ is equal to 40 multiplied by π·πΆ plus 31. Distributing the parentheses on the
right, we have 40π·πΆ plus 1240. And now subtracting 40π·πΆ from
both sides, we have 20π·πΆ is 1240. Dividing both sides by 20, we have
π·πΆ is 62. And since π·πΆ is the same as πΆπ·,
we have πΆπ· is equal to 62 units.

Up till now, weβve discussed the
relationship between the ratio of lengths of line segments related to the bisector
of an interior or exterior angle of a triangle. Letβs consider now a theorem that
deals with the length of the angle bisector line segment.

Given a triangle π΄π΅πΆ, if the
segment πΆπ· is the angle bisector of angle πΆ, then we have the length πΆπ· is the
square root of πΆπ΅ multiplied by πΆπ΄ minus π·π΅ multiplied by π·π΄. We use this theorem to work out the
length of the angle bisector within a triangle. To prove this theorem, we begin by
adding a circle circumscribing the triangle and also adding the point πΈ obtained by
extending the line πΆπ· to the circle.

In the diagram, we know that the
two green angles at the vertex πΆ are congruent since the segment πΆπ· is the
bisector of the angle π΄πΆπ΅. We also note that the angles at the
vertices π΅ and πΈ are inscribed angles subtended by the common arc π΄πΆ. And we recall that all angles
subtended by the common arc have equal measurements. This means that the angles at π΅
and πΈ are congruent. We then have the triangle πΆπ΅π·
and triangle πΆπΈπ΄ share two pairs of congruent angles. Sharing two pairs of congruent
angles means that the triangles are similar triangles. And this in turn means that the
ratio πΆπ· to πΆπ΄ is the same as that of πΆπ΅ to πΆπΈ.

Now, we know that πΆπΈ is equal to
πΆπ· plus π·πΈ. And we can insert this into the
denominator on our right-hand side. Then multiplying both sides by πΆπ΄
and πΆπ· plus π·πΈ and distributing our parentheses, we can then subtract πΆπ·
multiplied by π·πΈ from both sides. If we took the square root on both
sides, we wouldnβt quite yet have our formula. We need to replace the term πΆπ·
multiplied by π·πΈ with π·π΅ multiplied by π·π΄. And to justify this substitution,
we need to observe another pair of similar triangles.

In our diagram, we can see that the
angle π΅πΆπΈ and the angle π΅π΄πΈ are inscribed angles subtended by the same
arc. And this means that theyβre
congruent. Now this means the triangle πΆπ΅π·
and triangle π΄πΈπ· share two pairs of congruent angles. And this means that theyβre similar
triangles. We therefore have that the ratio
πΆπ· to π·π΄ is the same as π·π΅ to π·πΈ. And now if we multiply both sides
by π·π΄ and π·πΈ, weβre left with π·πΈ multiplied by πΆπ· is equal to π·π΅
multiplied by π·π΄. And we can now replace πΆπ·
multiplied by π·πΈ with π·π΅ multiplied by π·π΄. And now taking the positive square
root on both sides, we have πΆπ· is equal to the square root of πΆπ΅ multiplied by
πΆπ΄ minus π·π΅ multiplied by π·π΄, which is what we required and therefore proves
the theorem.

Now letβs look at an example where
we apply this theorem to find the length of the bisector of an interior angle of a
triangle.

In the triangle π΄π΅πΆ, π΄π΅ is 76
centimeters, π΄πΆ is 57 centimeters, and π΅π· is 52 centimeters. Given that the line segment π΄π·
bisects the angle π΄ and intersects the line segment π΅πΆ at π·, determine the
length of the line segment π΄π·.

We recall that if the line segment
π΄π· bisects the angle π΄ in a triangle, then we have the theorem π΄π· is equal to
the square root of π΄πΆ multiplied by π΄π΅ minus π·πΆ multiplied by π·π΅. Now from the question, we know that
π΄πΆ is equal to 57, π΅π·, which is π·π΅, is equal to 52, and that π΄π΅ has length
76. And so to find the length π΄π·, we
must first find the length π·πΆ. To do this, weβre going to use the
interior angle bisector theorem. This says that the bisector of an
interior angle of a triangle divides the opposite side into segments whose lengths
have the same ratio as the lengths of the noncommon adjacent sides of the bisected
angle.

What this means in our triangle is
that π·πΆ over π΅π· is the same as π΄πΆ over π΄π΅. Now we know that π΅π· is 52, π΄πΆ
is 57, and π΄π΅ is 76. So we have π·πΆ over 52 is equal to
57 over 76. Now, multiplying both sides by 52,
weβre left with π·πΆ on the left-hand side. And 57 divided by 76 multiplied by
52 is 39. So π·πΆ is 39 centimeters. We now have all the information we
need to calculate π΄π·. Substituting in our values, we have
π΄π· is the square root of 57 times 76 minus 39 times 52, that is, the square root
of 4332 minus 2028, which is the square root of 2304, that is, 48. The length of the line segment π΄π·
is therefore 48 centimeters.

Letβs finish by recapping some of
the important concepts weβve covered. You know that 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 bisected angle. That is, from the diagram shown,
the ratio π·π΅ over π·π΄ is the same as πΆπ΅ over πΆπ΄. The converse is also true where if
weβre given a point π· on the side π΅πΆ and given that π·π΅ over π·πΆ is π΄π΅ over
π΄πΆ, then the point π· lies on the bisector of the interior angle π΄.

Next, if we consider a bisector of
an exterior angle at vertex πΆ of a triangle π΄π΅πΆ, which intersects an extension
of the side of the triangle opposite to the angle at the point π·, then the ratio of
the length π·π΅ to π·π΄ is the same as the ratio of the length πΆπ΅ to πΆπ΄. And finally, in any triangle
π΄π΅πΆ, if the line segment πΆπ· is the angle bisector of angle πΆ, then we have
πΆπ· is equal to the square root of πΆπ΅ multiplied by πΆπ΄ minus π·π΅ multiplied by
π·π΄.