# 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.

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### Video Transcript

In this video, weβ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. Weβll consider this with some examples together with the exterior angle bisector theorem. And weβ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 π·π΅ divided by the length π·π΄ is the same as the length πΆπ΅ over the length πΆπ΄.

To prove this, letβs make some room and redraw our triangle. We begin by adding a new line from the vertex π΄ parallel to the line πΆπ·. We then extend the line segment π΅πΆ until it meets this new line. And the two lines intersect at the point πΈ. Since the line πΆπ· is the bisector of the angle at the point πΆ, we know that the two angles in the diagram at vertex πΆ are congruent. That is, theyβre the same. Now, since line segments π΄πΈ and πΆπ· are parallel, then angles π΅πΆπ· and π΅πΈπ΄ are congruent. And where angles π΅πΆπ· and π΅πΈπ΄ are corresponding angles, angles π·πΆπ΄ and πΆπ΄πΈ are alternate angles and therefore also congruent. And this means that all four marked angles in the diagram are congruent.

And since triangles π΅πΆπ· and π΅πΈπ΄ share two congruent angles, theyβre similar triangles, then it follows that the ratio π΅π· over π΅π΄ is the same as the ratio π΅πΆ over π΅πΈ. And we could rewrite this as shown since π΅π΄ is the same as π΅π· plus π·π΄ and π΅πΈ is the same as π΅πΆ plus πΆπΈ. If we then cross multiply and distribute the parentheses, we have π΅π· multiplied by π΅πΆ plus π΅π· multiplied by πΆπΈ is equal to π΅πΆ multiplied by π΅π· plus π΅πΆ multiplied by π·π΄. Subtracting π΅π· multiplied by π΅πΆ from both sides, this then reduces to π΅π· multiplied by πΆπΈ is π΅πΆ multiplied by π·π΄. And now dividing both sides by π·π΄ multiplied by πΆπΈ, we can divide the numerator and denominator on the left by πΆπΈ 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΅πΆ divided by πΆπΈ.

Finally, we notice that the triangle π΄πΆπΈ is an isosceles triangle. And this means that πΆπΈ is equal to πΆπ΄. And we can replace this in our equation. So we have π΅π· over π·π΄ is equal to π΅πΆ over πΆπ΄. And simply rewriting π΅π· as π·π΅ and π΅πΆ as πΆπ΅, we have a formula π·π΅ over π·π΄ is equal to πΆπ΅ over πΆπ΄. Letβs 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΄πΆ, π΅π· is equal to eight, π·πΆ is equal to 11, and the perimeter of the triangle π΄π΅πΆ is 57. Determine the lengths of the line segments π΄π΅ and π΄πΆ.

Weβre told that the line segment π΄π· bisects the angle at π΄. And to answer this question, weβ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 π·πΆ divided by the length π΅π· is the same as the length π΄πΆ divided by the length π΄π΅.

Weβre given that π΅π· is equal to eight and π·πΆ is equal to 11. And this means that 11 over eight is equal to π΄πΆ over π΄π΅. Weβ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 π΄π΅ plus π΅πΆ plus π΄πΆ is 57. And since we know that π΅πΆ is πΆπ· plus π·π΅, and thatβs 11 plus eight, we can replace π΅πΆ with 19. Now, if we subtract 19 from both sides, we have π΄π΅ plus π΄πΆ is equal to 38. Now if we recall our equation from earlier, thatβs 11 over eight is equal to π΄πΆ over π΄π΅, we can find π΄πΆ in terms of π΄π΅ by multiplying both sides by π΄π΅. This gives us 11 over eight π΄π΅ is equal to π΄πΆ.

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΄πΆ so that we have 11 over eight multiplied by 16 is equal to π΄πΆ. 11 over eight multiplied by 16 is 22. Therefore, π΄πΆ is 22. Hence, the length of the line segment π΄πΆ 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΅πΆ where weβre given a point π· on the side π΅πΆ. If weβre given also that the lengths of the line segments satisfy the equation π·π΅ over π·πΆ is equal to π΄π΅ over π΄πΆ, then we know that the point π· lies on the bisector of the interior angle π΅π΄πΆ. Now, letβs look at an analogous bisector theorem for an exterior angle of a triangle.

Consider the bisector of an exterior angle at vertex πΆ of a triangle π΄π΅πΆ 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 πΆπ΅ over πΆπ΄. That is, the ratio of the length π·π΅ to the length π·π΄ is equal to the ratio of the length πΆπ΅ to the length πΆπ΄. And this is the exterior angle bisector theorem. Although we wonβt prove this here, we know that the ideas are similar, though more complex, to those used in proving the interior angle bisector theorem. Letβs 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΄πΆ is 40, and π΅πΆ is 31, what is πΆπ·?

The diagram indicates that the two angles shown are congruent. That is, theyβre 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΅πΆ. We recall that the exterior angle bisector theorem gives us the identity π·πΆ over π·π΅ is equal to π΄πΆ over π΄π΅, that is, that the ratio of the length π·πΆ to the length π·π΅ is the same as the ratio of the length π΄πΆ to the length π΄π΅. Now, we know that the length π·π΅ is equal to π·πΆ plus πΆπ΅. So we can replace this in our formula. And we have π·πΆ over π·πΆ plus πΆπ΅ is equal to π΄πΆ over π΄π΅.

Weβre given the length π΄π΅ is equal to 60, π΅πΆ is equal to 31, and π΄πΆ is equal to 40. Into our equation then, the equation has only one unknown; thatβs π·πΆ. And our equation is then π·πΆ divided by π·πΆ plus 31 is 40 over 60. And notice since the length π·πΆ is the same as πΆπ·, this is what weβre looking for. So now letβs solve our equation for π·πΆ. Multiplying both sides by 60 and also π·πΆ plus 31, we have 60π·πΆ is equal to 40 multiplied by π·πΆ plus 31. Distributing the parentheses on the right, we have 40π·πΆ plus 1240. And now subtracting 40π·πΆ from both sides, we have 20π·πΆ is 1240. Dividing both sides by 20, we have π·πΆ is 62. And since π·πΆ is the same as πΆπ·, we have πΆπ· is equal to 62 units.

Up till now, weβ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. Letβs consider now a theorem that deals with the length of the angle bisector line segment.

Given a triangle π΄π΅πΆ, if the segment πΆπ· is the angle bisector of angle πΆ, then we have the length πΆπ· is the square root of πΆπ΅ multiplied by πΆπ΄ 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 πΈ obtained by extending the line πΆπ· to the circle.

In the diagram, we know that the two green angles at the vertex πΆ are congruent since the segment πΆπ· is the bisector of the angle π΄πΆπ΅. We also note that the angles at the vertices π΅ and πΈ are inscribed angles subtended by the common arc π΄πΆ. And we recall that all angles subtended by the common arc have equal measurements. This means that the angles at π΅ and πΈ are congruent. We then have the triangle πΆπ΅π· and triangle πΆπΈπ΄ 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 πΆπ· to πΆπ΄ is the same as that of πΆπ΅ to πΆπΈ.

Now, we know that πΆπΈ is equal to πΆπ· plus π·πΈ. And we can insert this into the denominator on our right-hand side. Then multiplying both sides by πΆπ΄ and πΆπ· plus π·πΈ and distributing our parentheses, we can then subtract πΆπ· multiplied by π·πΈ from both sides. If we took the square root on both sides, we wouldnβt quite yet have our formula. We need to replace the term πΆπ· multiplied by π·πΈ 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΅πΆπΈ and the angle π΅π΄πΈ are inscribed angles subtended by the same arc. And this means that theyβre congruent. Now this means the triangle πΆπ΅π· and triangle π΄πΈπ· share two pairs of congruent angles. And this means that theyβre similar triangles. We therefore have that the ratio πΆπ· to π·π΄ is the same as π·π΅ to π·πΈ. And now if we multiply both sides by π·π΄ and π·πΈ, weβre left with π·πΈ multiplied by πΆπ· is equal to π·π΅ multiplied by π·π΄. And we can now replace πΆπ· multiplied by π·πΈ with π·π΅ multiplied by π·π΄. And now taking the positive square root on both sides, we have πΆπ· is equal to the square root of πΆπ΅ multiplied by πΆπ΄ minus π·π΅ multiplied by π·π΄, which is what we required and therefore proves the theorem.

Now letβ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 π΄π΅πΆ, π΄π΅ is 76 centimeters, π΄πΆ 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΅πΆ 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΄πΆ multiplied by π΄π΅ minus π·πΆ multiplied by π·π΅. Now from the question, we know that π΄πΆ 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·πΆ. To do this, weβ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 π·πΆ over π΅π· is the same as π΄πΆ over π΄π΅. Now we know that π΅π· is 52, π΄πΆ is 57, and π΄π΅ is 76. So we have π·πΆ over 52 is equal to 57 over 76. Now, multiplying both sides by 52, weβre left with π·πΆ on the left-hand side. And 57 divided by 76 multiplied by 52 is 39. So π·πΆ 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.

Letβs finish by recapping some of the important concepts weβ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 π·π΅ over π·π΄ is the same as πΆπ΅ over πΆπ΄. The converse is also true where if weβre given a point π· on the side π΅πΆ and given that π·π΅ over π·πΆ is π΄π΅ over π΄πΆ, 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 πΆ of a triangle π΄π΅πΆ, 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 πΆπ΅ to πΆπ΄. And finally, in any triangle π΄π΅πΆ, if the line segment πΆπ· is the angle bisector of angle πΆ, then we have πΆπ· is equal to the square root of πΆπ΅ multiplied by πΆπ΄ minus π·π΅ multiplied by π·π΄.