Difference between revisions of "Angle Bisector Theorem"

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== Proof ==
 
== Proof ==
  
There is currently no text here. Help us make a proof.
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By LoS on ACD and ABD,
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AB/BD=sin(BDA)/sin(BAD) ... (1)
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AC/AD=sin(ADC)/sin(DAC) ... (2)
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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
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sin(BDA)=sin(ADC)
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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 13:49, 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, AB/BD=sin(BDA)/sin(BAD) ... (1) 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.

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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