Difference between revisions of "Angle Bisector Theorem"
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=== Method 3 === | === Method 3 === | ||
+ | Let <math>AD = d</math>. Now, we can express the area of triangle ABD in two ways: | ||
− | + | <math>[ABD] = \frac 12 cd\sin \angle BAD = \frac 12 md \sin \angle ADB.</math> | |
+ | Thus, <math>\frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm</math>. | ||
+ | |||
+ | Likewise, triangle ACD can be expressed in two different ways: | ||
+ | |||
+ | <math>[ACD] = \frac 12 bd \sin \angle CAD = \frac 12 dn \sin \angle ADC.</math> | ||
+ | Thus, <math>\frac{\sin \angle ADC}{\sin \angle CAD} = \frac bn</math>. | ||
+ | |||
+ | But <math>\angle CAD \cong \angle BAD</math> and <math>\sin \angle ADC = \sin \angle ADB</math> since <math>\angle ADC = \pi - \angle ADB</math>. Therefore, we can substitute back into our previous equation to get <math>\frac{\sin \angle ADB}{\sin \angle BAD} = \frac bn</math>. | ||
+ | |||
+ | We conclude that <math>\frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm = \frac bn</math>, which was what we wanted. | ||
+ | |||
+ | In both cases, if we reverse all the steps, we see that everything still holds and thus the converse holds. | ||
== Examples == | == Examples == |
Revision as of 10:47, 17 April 2015
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Introduction
The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then
. Likewise, the converse of this theorem holds as well.
Proof
Method 1
Because of the ratios and equal angles in the theorem, we think of similar triangles. There are not any similar triangles in the figure as it now stands, however. So, we think to draw in a carefully chosen line or two. Extending AD until it hits the line through C parallel to AB does just the trick:
Since AB and CE are parallel, we know that and
. Triangle ACE is isosceles, with AC = CE.
By AA similarity, . By the properties of similar triangles, we arrive at our desired result:
![$\frac cm = \frac bn.$](http://latex.artofproblemsolving.com/c/f/1/cf1574360a930676f4336884fe3bed4fc7d32024.png)
Method 2
Let . Now, we can express the area of triangle ABD in two ways:
![$[ABD] = \frac 12 cd\sin \angle BAD = \frac 12 md \sin \angle ADB.$](http://latex.artofproblemsolving.com/9/1/e/91ed453f1f6383386b7e7c2316f952e2d76c6f93.png)
Thus, .
Likewise, triangle ACD can be expressed in two different ways:
![$[ACD] = \frac 12 bd \sin \angle CAD = \frac 12 dn \sin \angle ADC.$](http://latex.artofproblemsolving.com/6/1/9/6194b407398cbb9835d9512e991c68bfc3aaac74.png)
Thus, .
But and
since
. Therefore, we can substitute back into our previous equation to get
.
We conclude that , which was what we wanted.
In both cases, if we reverse all the steps, we see that everything still holds and thus the converse holds.
Method 3
Let . Now, we can express the area of triangle ABD in two ways:
Thus,
.
Likewise, triangle ACD can be expressed in two different ways:
Thus,
.
But and
since
. Therefore, we can substitute back into our previous equation to get
.
We conclude that , which was what we wanted.
In both cases, if we reverse all the steps, we see that everything still holds and thus the converse holds.
Examples
- Let ABC be a triangle with angle bisector AD with D on line segment BC. If
and
, find AB and AC.
Solution: By the angle bisector theorem,or
. Plugging this into
and solving for AC gives
. We can plug this back in to find
.
- In triangle ABC, let P be a point on BC and let
. Find the value of
.
Solution: First, we notice that. Thus, AP is the angle bisector of angle A, making our answer 0.
- Part (b), 1959 IMO Problems/Problem 5.