Difference between revisions of "Angle Bisector Theorem"
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By the [[Law of Sines]] on <math>\angle ACD</math> and <math>\angle ABD</math>, | By the [[Law of Sines]] on <math>\angle ACD</math> and <math>\angle ABD</math>, | ||
− | \begin{align*}\frac{AB}{BD}&=\frac{\sin(BDA)}{\sin(BAD)}\\ | + | <cmath>\begin{align*}\frac{AB}{BD}&=\frac{\sin(BDA)}{\sin(BAD)}\\ |
− | \frac{AC}{AD}&=\frac{\sin(ADC)}{\sin(CAD)}\end{align*} | + | \frac{AC}{AD}&=\frac{\sin(ADC)}{\sin(CAD)}\end{align*}</cmath> |
First, because <math>\bar{AD}</math> is an angle bisector, we know that <math>m\angle BAD = m\angle CAD</math> and thus <math>\sin(BAD) = \sin(CAD)</math>, so the denominators are equal. | First, because <math>\bar{AD}</math> is an angle bisector, we know that <math>m\angle BAD = m\angle CAD</math> and thus <math>\sin(BAD) = \sin(CAD)</math>, so the denominators are equal. |
Revision as of 23:11, 12 June 2020
This is an AoPSWiki Word of the Week for June 6-12 |
Introduction & Formulas
The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then
. It follows that
. Likewise, the converse of this theorem holds as well.
Further by combining with Stewart's Theorem it can be shown that
Proof
By the Law of Sines on and
,
First, because is an angle bisector, we know that
and thus
, so the denominators are equal.
Second, we observe that and
.
Therefore,
, so the numerators are equal.
It then follows that
Examples & Problems
- 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.