# Lesson Video: Angle Bisector Theorem and Its Converse Mathematics

In this video, we will learn how to use the angle bisector theorem and its converse to find a missing side length in a triangle.

15:19

### Video Transcript

In this video, we’ll 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. We’ll consider this with some examples together with the exterior angle bisector theorem. And we’ll 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 𝐷𝐵 divided by the length 𝐷𝐴 is the same as the length 𝐶𝐵 over the length 𝐶𝐴.

To prove this, let’s make some room and redraw our triangle. We begin by adding a new line from the vertex 𝐴 parallel to the line 𝐶𝐷. We then extend the line segment 𝐵𝐶 until it meets this new line. And the two lines intersect at the point 𝐸. Since the line 𝐶𝐷 is the bisector of the angle at the point 𝐶, we know that the two angles in the diagram at vertex 𝐶 are congruent. That is, they’re the same. Now, since line segments 𝐴𝐸 and 𝐶𝐷 are parallel, then angles 𝐵𝐶𝐷 and 𝐵𝐸𝐴 are congruent. And where angles 𝐵𝐶𝐷 and 𝐵𝐸𝐴 are corresponding angles, angles 𝐷𝐶𝐴 and 𝐶𝐴𝐸 are alternate angles and therefore also congruent. And this means that all four marked angles in the diagram are congruent.

And since triangles 𝐵𝐶𝐷 and 𝐵𝐸𝐴 share two congruent angles, they’re similar triangles, then it follows that the ratio 𝐵𝐷 over 𝐵𝐴 is the same as the ratio 𝐵𝐶 over 𝐵𝐸. And we could rewrite this as shown since 𝐵𝐴 is the same as 𝐵𝐷 plus 𝐷𝐴 and 𝐵𝐸 is the same as 𝐵𝐶 plus 𝐶𝐸. If we then cross multiply and distribute the parentheses, we have 𝐵𝐷 multiplied by 𝐵𝐶 plus 𝐵𝐷 multiplied by 𝐶𝐸 is equal to 𝐵𝐶 multiplied by 𝐵𝐷 plus 𝐵𝐶 multiplied by 𝐷𝐴. Subtracting 𝐵𝐷 multiplied by 𝐵𝐶 from both sides, this then reduces to 𝐵𝐷 multiplied by 𝐶𝐸 is 𝐵𝐶 multiplied by 𝐷𝐴. And now dividing both sides by 𝐷𝐴 multiplied by 𝐶𝐸, we can divide the numerator and denominator on the left by 𝐶𝐸 and the numerator and denominator on the right by 𝐷𝐴. And making some room, we have 𝐵𝐷 divided by 𝐷𝐴 is equal to 𝐵𝐶 divided by 𝐶𝐸.

Finally, we notice that the triangle 𝐴𝐶𝐸 is an isosceles triangle. And this means that 𝐶𝐸 is equal to 𝐶𝐴. And we can replace this in our equation. So we have 𝐵𝐷 over 𝐷𝐴 is equal to 𝐵𝐶 over 𝐶𝐴. And simply rewriting 𝐵𝐷 as 𝐷𝐵 and 𝐵𝐶 as 𝐶𝐵, we have a formula 𝐷𝐵 over 𝐷𝐴 is equal to 𝐶𝐵 over 𝐶𝐴. Let’s look at an example of how we use this theorem to find missing lengths in a triangle.

In the figure, the line segment 𝐴𝐷 bisects the angle 𝐵𝐴𝐶, 𝐵𝐷 is equal to eight, 𝐷𝐶 is equal to 11, and the perimeter of the triangle 𝐴𝐵𝐶 is 57. Determine the lengths of the line segments 𝐴𝐵 and 𝐴𝐶.

We’re told that the line segment 𝐴𝐷 bisects the angle at 𝐴. And to answer this question, we’re 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 𝐷𝐶 divided by the length 𝐵𝐷 is the same as the length 𝐴𝐶 divided by the length 𝐴𝐵.

We’re given that 𝐵𝐷 is equal to eight and 𝐷𝐶 is equal to 11. And this means that 11 over eight is equal to 𝐴𝐶 over 𝐴𝐵. We’re 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 𝐴𝐵 plus 𝐵𝐶 plus 𝐴𝐶 is 57. And since we know that 𝐵𝐶 is 𝐶𝐷 plus 𝐷𝐵, and that’s 11 plus eight, we can replace 𝐵𝐶 with 19. Now, if we subtract 19 from both sides, we have 𝐴𝐵 plus 𝐴𝐶 is equal to 38. Now if we recall our equation from earlier, that’s 11 over eight is equal to 𝐴𝐶 over 𝐴𝐵, we can find 𝐴𝐶 in terms of 𝐴𝐵 by multiplying both sides by 𝐴𝐵. This gives us 11 over eight 𝐴𝐵 is equal to 𝐴𝐶.

And now replacing this in our second equation, we have 𝐴𝐵 plus 11 over eight 𝐴𝐵 is equal to 38. Now collecting our terms on the left, we have 19 over eight 𝐴𝐵 is equal to 38. Now, multiplying both sides by eight and dividing by 19, we have 𝐴𝐵 is 38 multiplied by eight over 19 so that 𝐴𝐵 is equal to 16. We can now use this in our equation for 𝐴𝐶 so that we have 11 over eight multiplied by 16 is equal to 𝐴𝐶. 11 over eight multiplied by 16 is 22. Therefore, 𝐴𝐶 is 22. Hence, the length of the line segment 𝐴𝐶 is 22 and that of 𝐴𝐵 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 𝐴𝐵𝐶 where we’re given a point 𝐷 on the side 𝐵𝐶. If we’re given also that the lengths of the line segments satisfy the equation 𝐷𝐵 over 𝐷𝐶 is equal to 𝐴𝐵 over 𝐴𝐶, then we know that the point 𝐷 lies on the bisector of the interior angle 𝐵𝐴𝐶. Now, let’s look at an analogous bisector theorem for an exterior angle of a triangle.

Consider the bisector of an exterior angle at vertex 𝐶 of a triangle 𝐴𝐵𝐶 intersecting an extension of the side of the triangle opposite to the angle at a point 𝐷. We then have the identity 𝐷𝐵 over 𝐷𝐴 is equal to 𝐶𝐵 over 𝐶𝐴. That is, the ratio of the length 𝐷𝐵 to the length 𝐷𝐴 is equal to the ratio of the length 𝐶𝐵 to the length 𝐶𝐴. And this is the exterior angle bisector theorem. Although we won’t prove this here, we know that the ideas are similar, though more complex, to those used in proving the interior angle bisector theorem. Let’s consider an example of how to apply the exterior angle bisector theorem to find a missing length in a triangle.

Given that 𝐴𝐵 is equal to 60, 𝐴𝐶 is 40, and 𝐵𝐶 is 31, what is 𝐶𝐷?

The diagram indicates that the two angles shown are congruent. That is, they’re the same number of degrees or radians. The line segment 𝐴𝐷 is then the bisector of the exterior angle at 𝐴 of the triangle 𝐴𝐵𝐶. We recall that the exterior angle bisector theorem gives us the identity 𝐷𝐶 over 𝐷𝐵 is equal to 𝐴𝐶 over 𝐴𝐵, that is, that the ratio of the length 𝐷𝐶 to the length 𝐷𝐵 is the same as the ratio of the length 𝐴𝐶 to the length 𝐴𝐵. Now, we know that the length 𝐷𝐵 is equal to 𝐷𝐶 plus 𝐶𝐵. So we can replace this in our formula. And we have 𝐷𝐶 over 𝐷𝐶 plus 𝐶𝐵 is equal to 𝐴𝐶 over 𝐴𝐵.

We’re given the length 𝐴𝐵 is equal to 60, 𝐵𝐶 is equal to 31, and 𝐴𝐶 is equal to 40. Into our equation then, the equation has only one unknown; that’s 𝐷𝐶. And our equation is then 𝐷𝐶 divided by 𝐷𝐶 plus 31 is 40 over 60. And notice since the length 𝐷𝐶 is the same as 𝐶𝐷, this is what we’re looking for. So now let’s solve our equation for 𝐷𝐶. Multiplying both sides by 60 and also 𝐷𝐶 plus 31, we have 60𝐷𝐶 is equal to 40 multiplied by 𝐷𝐶 plus 31. Distributing the parentheses on the right, we have 40𝐷𝐶 plus 1240. And now subtracting 40𝐷𝐶 from both sides, we have 20𝐷𝐶 is 1240. Dividing both sides by 20, we have 𝐷𝐶 is 62. And since 𝐷𝐶 is the same as 𝐶𝐷, we have 𝐶𝐷 is equal to 62 units.

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

Given a triangle 𝐴𝐵𝐶, if the segment 𝐶𝐷 is the angle bisector of angle 𝐶, then we have the length 𝐶𝐷 is the square root of 𝐶𝐵 multiplied by 𝐶𝐴 minus 𝐷𝐵 multiplied by 𝐷𝐴. 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 𝐸 obtained by extending the line 𝐶𝐷 to the circle.

In the diagram, we know that the two green angles at the vertex 𝐶 are congruent since the segment 𝐶𝐷 is the bisector of the angle 𝐴𝐶𝐵. We also note that the angles at the vertices 𝐵 and 𝐸 are inscribed angles subtended by the common arc 𝐴𝐶. And we recall that all angles subtended by the common arc have equal measurements. This means that the angles at 𝐵 and 𝐸 are congruent. We then have the triangle 𝐶𝐵𝐷 and triangle 𝐶𝐸𝐴 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 𝐶𝐷 to 𝐶𝐴 is the same as that of 𝐶𝐵 to 𝐶𝐸.

Now, we know that 𝐶𝐸 is equal to 𝐶𝐷 plus 𝐷𝐸. And we can insert this into the denominator on our right-hand side. Then multiplying both sides by 𝐶𝐴 and 𝐶𝐷 plus 𝐷𝐸 and distributing our parentheses, we can then subtract 𝐶𝐷 multiplied by 𝐷𝐸 from both sides. If we took the square root on both sides, we wouldn’t quite yet have our formula. We need to replace the term 𝐶𝐷 multiplied by 𝐷𝐸 with 𝐷𝐵 multiplied by 𝐷𝐴. And to justify this substitution, we need to observe another pair of similar triangles.

In our diagram, we can see that the angle 𝐵𝐶𝐸 and the angle 𝐵𝐴𝐸 are inscribed angles subtended by the same arc. And this means that they’re congruent. Now this means the triangle 𝐶𝐵𝐷 and triangle 𝐴𝐸𝐷 share two pairs of congruent angles. And this means that they’re similar triangles. We therefore have that the ratio 𝐶𝐷 to 𝐷𝐴 is the same as 𝐷𝐵 to 𝐷𝐸. And now if we multiply both sides by 𝐷𝐴 and 𝐷𝐸, we’re left with 𝐷𝐸 multiplied by 𝐶𝐷 is equal to 𝐷𝐵 multiplied by 𝐷𝐴. And we can now replace 𝐶𝐷 multiplied by 𝐷𝐸 with 𝐷𝐵 multiplied by 𝐷𝐴. And now taking the positive square root on both sides, we have 𝐶𝐷 is equal to the square root of 𝐶𝐵 multiplied by 𝐶𝐴 minus 𝐷𝐵 multiplied by 𝐷𝐴, which is what we required and therefore proves the theorem.

Now let’s 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 𝐴𝐵𝐶, 𝐴𝐵 is 76 centimeters, 𝐴𝐶 is 57 centimeters, and 𝐵𝐷 is 52 centimeters. Given that the line segment 𝐴𝐷 bisects the angle 𝐴 and intersects the line segment 𝐵𝐶 at 𝐷, determine the length of the line segment 𝐴𝐷.

We recall that if the line segment 𝐴𝐷 bisects the angle 𝐴 in a triangle, then we have the theorem 𝐴𝐷 is equal to the square root of 𝐴𝐶 multiplied by 𝐴𝐵 minus 𝐷𝐶 multiplied by 𝐷𝐵. Now from the question, we know that 𝐴𝐶 is equal to 57, 𝐵𝐷, which is 𝐷𝐵, is equal to 52, and that 𝐴𝐵 has length 76. And so to find the length 𝐴𝐷, we must first find the length 𝐷𝐶. To do this, we’re 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 𝐷𝐶 over 𝐵𝐷 is the same as 𝐴𝐶 over 𝐴𝐵. Now we know that 𝐵𝐷 is 52, 𝐴𝐶 is 57, and 𝐴𝐵 is 76. So we have 𝐷𝐶 over 52 is equal to 57 over 76. Now, multiplying both sides by 52, we’re left with 𝐷𝐶 on the left-hand side. And 57 divided by 76 multiplied by 52 is 39. So 𝐷𝐶 is 39 centimeters. We now have all the information we need to calculate 𝐴𝐷. Substituting in our values, we have 𝐴𝐷 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 𝐴𝐷 is therefore 48 centimeters.

Let’s finish by recapping some of the important concepts we’ve 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 𝐷𝐵 over 𝐷𝐴 is the same as 𝐶𝐵 over 𝐶𝐴. The converse is also true where if we’re given a point 𝐷 on the side 𝐵𝐶 and given that 𝐷𝐵 over 𝐷𝐶 is 𝐴𝐵 over 𝐴𝐶, then the point 𝐷 lies on the bisector of the interior angle 𝐴.

Next, if we consider a bisector of an exterior angle at vertex 𝐶 of a triangle 𝐴𝐵𝐶, which intersects an extension of the side of the triangle opposite to the angle at the point 𝐷, then the ratio of the length 𝐷𝐵 to 𝐷𝐴 is the same as the ratio of the length 𝐶𝐵 to 𝐶𝐴. And finally, in any triangle 𝐴𝐵𝐶, if the line segment 𝐶𝐷 is the angle bisector of angle 𝐶, then we have 𝐶𝐷 is equal to the square root of 𝐶𝐵 multiplied by 𝐶𝐴 minus 𝐷𝐵 multiplied by 𝐷𝐴.