Difference between revisions of "Angle Bisector Theorem"
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== Introduction == | == Introduction == | ||
The '''Angle Bisector Theorem''' states that given [[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC, then <math> \frac cm = \frac bn </math>. It follows that <math> \frac cb = \frac mn </math>. Likewise, the [[converse]] of this theorem holds as well. | The '''Angle Bisector Theorem''' states that given [[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC, then <math> \frac cm = \frac bn </math>. It follows that <math> \frac cb = \frac mn </math>. Likewise, the [[converse]] of this theorem holds as well. | ||
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| + | Further by combining with [[Stewart's Theorem]] it can be shown that <math>AD^2 = b\cdot c - m \cdot n</math> | ||
<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> | <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> | ||
Revision as of 20:23, 15 February 2019
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Introduction
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)
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.
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.
