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 𝐷𝐡 divided by the length 𝐷𝐴 is the same as the length 𝐢𝐡 over the length 𝐢𝐴.

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 𝐢𝐷. We then extend the line segment 𝐡𝐢 until it meets this new line. And the two lines intersect at the point 𝐸. Since the line 𝐢𝐷 is the bisector of the angle at the point 𝐢, we know that the two angles in the diagram at vertex 𝐢 are congruent. That is, they’re the same. Now, since line segments 𝐴𝐸 and 𝐢𝐷 are parallel, then angles 𝐡𝐢𝐷 and 𝐡𝐸𝐴 are congruent. And where angles 𝐡𝐢𝐷 and 𝐡𝐸𝐴 are corresponding angles, angles 𝐷𝐢𝐴 and 𝐢𝐴𝐸 are alternate angles and therefore also congruent. And this means that all four marked angles in the diagram are congruent.

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

Finally, we notice that the triangle 𝐴𝐢𝐸 is an isosceles triangle. And this means that 𝐢𝐸 is equal to 𝐢𝐴. And we can replace this in our equation. So we have 𝐡𝐷 over 𝐷𝐴 is equal to 𝐡𝐢 over 𝐢𝐴. And simply rewriting 𝐡𝐷 as 𝐷𝐡 and 𝐡𝐢 as 𝐢𝐡, we have a formula 𝐷𝐡 over 𝐷𝐴 is equal to 𝐢𝐡 over 𝐢𝐴. 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 𝐡𝐴𝐢, 𝐡𝐷 is equal to eight, 𝐷𝐢 is equal to 11, and the perimeter of the triangle 𝐴𝐡𝐢 is 57. Determine the lengths of the line segments 𝐴𝐡 and 𝐴𝐢.

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 𝐷𝐢 divided by the length 𝐡𝐷 is the same as the length 𝐴𝐢 divided by the length 𝐴𝐡.

We’re given that 𝐡𝐷 is equal to eight and 𝐷𝐢 is equal to 11. And this means that 11 over eight is equal to 𝐴𝐢 over 𝐴𝐡. 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 𝐴𝐡 plus 𝐡𝐢 plus 𝐴𝐢 is 57. And since we know that 𝐡𝐢 is 𝐢𝐷 plus 𝐷𝐡, and that’s 11 plus eight, we can replace 𝐡𝐢 with 19. Now, if we subtract 19 from both sides, we have 𝐴𝐡 plus 𝐴𝐢 is equal to 38. Now if we recall our equation from earlier, that’s 11 over eight is equal to 𝐴𝐢 over 𝐴𝐡, we can find 𝐴𝐢 in terms of 𝐴𝐡 by multiplying both sides by 𝐴𝐡. This gives us 11 over eight 𝐴𝐡 is equal to 𝐴𝐢.

And now replacing this in our second equation, we have 𝐴𝐡 plus 11 over eight 𝐴𝐡 is equal to 38. Now collecting our terms on the left, we have 19 over eight 𝐴𝐡 is equal to 38. Now, multiplying both sides by eight and dividing by 19, we have 𝐴𝐡 is 38 multiplied by eight over 19 so that 𝐴𝐡 is equal to 16. We can now use this in our equation for 𝐴𝐢 so that we have 11 over eight multiplied by 16 is equal to 𝐴𝐢. 11 over eight multiplied by 16 is 22. Therefore, 𝐴𝐢 is 22. Hence, the length of the line segment 𝐴𝐢 is 22 and that of 𝐴𝐡 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 𝐴𝐡𝐢 where we’re given a point 𝐷 on the side 𝐡𝐢. If we’re given also that the lengths of the line segments satisfy the equation 𝐷𝐡 over 𝐷𝐢 is equal to 𝐴𝐡 over 𝐴𝐢, then we know that the point 𝐷 lies on the bisector of the interior angle 𝐡𝐴𝐢. 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 𝐢 of a triangle 𝐴𝐡𝐢 intersecting an extension of the side of the triangle opposite to the angle at a point 𝐷. We then have the identity 𝐷𝐡 over 𝐷𝐴 is equal to 𝐢𝐡 over 𝐢𝐴. That is, the ratio of the length 𝐷𝐡 to the length 𝐷𝐴 is equal to the ratio of the length 𝐢𝐡 to the length 𝐢𝐴. 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 𝐴𝐡 is equal to 60, 𝐴𝐢 is 40, and 𝐡𝐢 is 31, what is 𝐢𝐷?

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 𝐴𝐡𝐢. We recall that the exterior angle bisector theorem gives us the identity 𝐷𝐢 over 𝐷𝐡 is equal to 𝐴𝐢 over 𝐴𝐡, that is, that the ratio of the length 𝐷𝐢 to the length 𝐷𝐡 is the same as the ratio of the length 𝐴𝐢 to the length 𝐴𝐡. Now, we know that the length 𝐷𝐡 is equal to 𝐷𝐢 plus 𝐢𝐡. So we can replace this in our formula. And we have 𝐷𝐢 over 𝐷𝐢 plus 𝐢𝐡 is equal to 𝐴𝐢 over 𝐴𝐡.

We’re given the length 𝐴𝐡 is equal to 60, 𝐡𝐢 is equal to 31, and 𝐴𝐢 is equal to 40. Into our equation then, the equation has only one unknown; that’s 𝐷𝐢. And our equation is then 𝐷𝐢 divided by 𝐷𝐢 plus 31 is 40 over 60. And notice since the length 𝐷𝐢 is the same as 𝐢𝐷, this is what we’re looking for. So now let’s solve our equation for 𝐷𝐢. Multiplying both sides by 60 and also 𝐷𝐢 plus 31, we have 60𝐷𝐢 is equal to 40 multiplied by 𝐷𝐢 plus 31. Distributing the parentheses on the right, we have 40𝐷𝐢 plus 1240. And now subtracting 40𝐷𝐢 from both sides, we have 20𝐷𝐢 is 1240. Dividing both sides by 20, we have 𝐷𝐢 is 62. And since 𝐷𝐢 is the same as 𝐢𝐷, we have 𝐢𝐷 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 𝐴𝐡𝐢, if the segment 𝐢𝐷 is the angle bisector of angle 𝐢, then we have the length 𝐢𝐷 is the square root of 𝐢𝐡 multiplied by 𝐢𝐴 minus 𝐷𝐡 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 𝐢𝐷 to the circle.

In the diagram, we know that the two green angles at the vertex 𝐢 are congruent since the segment 𝐢𝐷 is the bisector of the angle 𝐴𝐢𝐡. We also note that the angles at the vertices 𝐡 and 𝐸 are inscribed angles subtended by the common arc 𝐴𝐢. And we recall that all angles subtended by the common arc have equal measurements. This means that the angles at 𝐡 and 𝐸 are congruent. We then have the triangle 𝐢𝐡𝐷 and triangle 𝐢𝐸𝐴 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 𝐢𝐷 to 𝐢𝐴 is the same as that of 𝐢𝐡 to 𝐢𝐸.

Now, we know that 𝐢𝐸 is equal to 𝐢𝐷 plus 𝐷𝐸. And we can insert this into the denominator on our right-hand side. Then multiplying both sides by 𝐢𝐴 and 𝐢𝐷 plus 𝐷𝐸 and distributing our parentheses, we can then subtract 𝐢𝐷 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 𝐢𝐷 multiplied by 𝐷𝐸 with 𝐷𝐡 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 𝐡𝐢𝐸 and the angle 𝐡𝐴𝐸 are inscribed angles subtended by the same arc. And this means that they’re congruent. Now this means the triangle 𝐢𝐡𝐷 and triangle 𝐴𝐸𝐷 share two pairs of congruent angles. And this means that they’re similar triangles. We therefore have that the ratio 𝐢𝐷 to 𝐷𝐴 is the same as 𝐷𝐡 to 𝐷𝐸. And now if we multiply both sides by 𝐷𝐴 and 𝐷𝐸, we’re left with 𝐷𝐸 multiplied by 𝐢𝐷 is equal to 𝐷𝐡 multiplied by 𝐷𝐴. And we can now replace 𝐢𝐷 multiplied by 𝐷𝐸 with 𝐷𝐡 multiplied by 𝐷𝐴. And now taking the positive square root on both sides, we have 𝐢𝐷 is equal to the square root of 𝐢𝐡 multiplied by 𝐢𝐴 minus 𝐷𝐡 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 𝐴𝐡𝐢, 𝐴𝐡 is 76 centimeters, 𝐴𝐢 is 57 centimeters, and 𝐡𝐷 is 52 centimeters. Given that the line segment 𝐴𝐷 bisects the angle 𝐴 and intersects the line segment 𝐡𝐢 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 𝐴𝐢 multiplied by 𝐴𝐡 minus 𝐷𝐢 multiplied by 𝐷𝐡. Now from the question, we know that 𝐴𝐢 is equal to 57, 𝐡𝐷, which is 𝐷𝐡, is equal to 52, and that 𝐴𝐡 has length 76. And so to find the length 𝐴𝐷, we must first find the length 𝐷𝐢. 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 𝐷𝐢 over 𝐡𝐷 is the same as 𝐴𝐢 over 𝐴𝐡. Now we know that 𝐡𝐷 is 52, 𝐴𝐢 is 57, and 𝐴𝐡 is 76. So we have 𝐷𝐢 over 52 is equal to 57 over 76. Now, multiplying both sides by 52, we’re left with 𝐷𝐢 on the left-hand side. And 57 divided by 76 multiplied by 52 is 39. So 𝐷𝐢 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 𝐷𝐡 over 𝐷𝐴 is the same as 𝐢𝐡 over 𝐢𝐴. The converse is also true where if we’re given a point 𝐷 on the side 𝐡𝐢 and given that 𝐷𝐡 over 𝐷𝐢 is 𝐴𝐡 over 𝐴𝐢, 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 𝐢 of a triangle 𝐴𝐡𝐢, which intersects an extension of the side of the triangle opposite to the angle at the point 𝐷, then the ratio of the length 𝐷𝐡 to 𝐷𝐴 is the same as the ratio of the length 𝐢𝐡 to 𝐢𝐴. And finally, in any triangle 𝐴𝐡𝐢, if the line segment 𝐢𝐷 is the angle bisector of angle 𝐢, then we have 𝐢𝐷 is equal to the square root of 𝐢𝐡 multiplied by 𝐢𝐴 minus 𝐷𝐡 multiplied by 𝐷𝐴.

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