Difference between revisions of "Angle Bisector Theorem"

(Proof)
(Proof)
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== Proof ==
 
== Proof ==
  
By LoS on ACD and ABD,
+
By LoS on ACD and ABD, \\*
 +
\newline
 
AB/BD=sin(BDA)/sin(BAD) ... (1)
 
AB/BD=sin(BDA)/sin(BAD) ... (1)
 +
\newline
 
AC/AD=sin(ADC)/sin(DAC) ... (2)
 
AC/AD=sin(ADC)/sin(DAC) ... (2)
 
Well, we also know that BDA and ADC add to 180 degrees. I think that means that we can use sin(180-x)=sin(x) here. Doing so, we see that
 
Well, we also know that BDA and ADC add to 180 degrees. I think that means that we can use sin(180-x)=sin(x) here. Doing so, we see that

Revision as of 13:50, 24 April 2020

This is an AoPSWiki Word of the Week for June 6-12

Introduction

The Angle Bisector Theorem states that given triangle $\triangle ABC$ and angle bisector AD, where D is on side BC, then $\frac cm = \frac bn$. It follows that $\frac cb = \frac mn$. Likewise, the converse of this theorem holds as well.


Further by combining with Stewart's Theorem it can be shown that $AD^2 = b\cdot c - m \cdot n$

[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]

Proof

By LoS on ACD and ABD, \\* \newline AB/BD=sin(BDA)/sin(BAD) ... (1) \newline AC/AD=sin(ADC)/sin(DAC) ... (2) Well, we also know that BDA and ADC add to 180 degrees. I think that means that we can use sin(180-x)=sin(x) here. Doing so, we see that sin(BDA)=sin(ADC) 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. ~ sachi2019

Examples

  1. Let ABC be a triangle with angle bisector AD with D on line segment BC. If $BD = 2, CD = 5,$ and $AB + AC = 10$, find AB and AC.
    Solution: By the angle bisector theorem, $\frac{AB}2 = \frac{AC}5$ or $AB = \frac 25 AC$. Plugging this into $AB + AC = 10$ and solving for AC gives $AC = \frac{50}7$. We can plug this back in to find $AB = \frac{20}7$.
  2. In triangle ABC, let P be a point on BC and let $AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3$. Find the value of $m\angle BAP - m\angle CAP$.
    Solution: First, we notice that $\frac{AB}{BP}=\frac{AC}{CP}$. Thus, AP is the angle bisector of angle A, making our answer 0.
  3. Part (b), 1959 IMO Problems/Problem 5.

See also

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