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 22:11, 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.