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 11:27, 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 
![[asy] size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1)); dot(A^^B^^C^^D,blue);label("$b$",(A+C)/2,(1,0));label("$c$",(A+B)/2,(0,1));label("$m$",(B+D)/2,(0,-1));label("$n$",(D+C)/2,(0,-1)); [/asy]](http://latex.artofproblemsolving.com/6/b/d/6bd1126eca956359a278a408df0ce6680fd7e11c.png)
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 ![\[\frac{AB}{BD}=\frac{\sin(BDA)}{\sin(BAD)} = \frac{AC}{CD}\]](http://latex.artofproblemsolving.com/3/d/9/3d9e5888753763436fc1f6ee57c713d2073c87e4.png)
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.
and 
, find AB and AC.
or 
. Plugging this into 
. We can plug this back in to find 
.
. Find the value of 
. 
. Thus, AP is the angle bisector of angle A, making our answer 0.
