Difference between revisions of "Angle Bisector Theorem"
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== Proof == | == Proof == | ||
| − | By LoS on | + | By <math>LoS</math> on <math>\angleACD</math> and <math>\angleABD</math>, |
| − | AB | + | <math>\frac{AB}{BD}=\frac{sin(BDA)}{sin(BAD)}</math> ... <math>(1)</math> and |
| − | AC | + | <math>\frac{AC}{AD}=\frac{sin(ADC)/sin(DAC)} </math>... <math>(2)</math> |
| − | Well, we also know that BDA and ADC add to 180 | + | Well, we also know that <math>\angle BDA</math> and <math>\angle ADC</math> add to <math>180^\circ</math>. I think that means that we can use <math>sin(180-x)=sin(x)</math> here. Doing so, we see that <math>sin(BDA)=sin(ADC)</math> |
| − | sin(BDA)=sin(ADC) | + | I noticed that these are the numerators of <math>(1)</math> and <math>(2)</math> respectively. Since <math>\angle BAD</math> and <math>\angle DAC</math> are equal, then you get the equation for the bisector angle theorem. ~ SilverLightning59 |
| − | I noticed that these are the numerators of (1) and (2) respectively. Since BAD and DAC are equal, then you get the equation for the bisector angle theorem. ~ | ||
== Examples == | == Examples == | ||
Revision as of 03:17, 26 April 2020
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Contents
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)
Proof
By 
on $\angleACD$ (Error compiling LaTeX. Unknown error_msg) and $\angleABD$ (Error compiling LaTeX. Unknown error_msg),
... 
and
$\frac{AC}{AD}=\frac{sin(ADC)/sin(DAC)}$ (Error compiling LaTeX. Unknown error_msg)... 
Well, we also know that 
and 
add to 
. I think that means that we can use 
here. Doing so, we see that 
I noticed that these are the numerators of and respectively. Since 
and 
are equal, then you get the equation for the bisector angle theorem. ~ SilverLightning59
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.
