Difference between revisions of "Angle Bisector Theorem"
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<cmath>\begin{align*}\frac{AB}{BD}&=\frac{\sin(BDA)}{\sin(BAD)}\\ | <cmath>\begin{align*}\frac{AB}{BD}&=\frac{\sin(BDA)}{\sin(BAD)}\\ | ||
− | \frac{AC}{ | + | \frac{AC}{CD}&=\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. | ||
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Therefore, <math>\sin(BDA) = \sin(CDA)</math>, so the numerators are equal. | Therefore, <math>\sin(BDA) = \sin(CDA)</math>, so the numerators are equal. | ||
− | It then follows that <cmath>\frac{AB}{BD}=\frac{\sin(BDA)}{\sin(BAD)} = \frac{AC}{ | + | It then follows that <cmath>\frac{AB}{BD}=\frac{\sin(BDA)}{\sin(BAD)} = \frac{AC}{CD}</cmath> |
== Examples & Problems == | == Examples & Problems == |
Revision as of 10:27, 21 December 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.