# Angle bisector theorem

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

## Introduction & Formulas

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 the Law of Sines on $\angle ACD$ and $\angle ABD$,

\begin{align*}\frac{AB}{BD}&=\frac{\sin(BDA)}{\sin(BAD)}\\ \frac{AC}{CD}&=\frac{\sin(ADC)}{\sin(CAD)}\end{align*}

First, because $\bar{AD}$ is an angle bisector, we know that $m\angle BAD = m\angle CAD$ and thus $\sin(BAD) = \sin(CAD)$, so the denominators are equal.

Second, we observe that $m\angle BDA + m\angle CDA = \pi$ and $\sin(\pi - \theta) = \sin(\theta)$. Therefore, $\sin(BDA) = \sin(CDA)$, so the numerators are equal.

It then follows that $$\frac{AB}{BD}=\frac{\sin(BDA)}{\sin(BAD)} = \frac{AC}{CD}$$

## Examples & Problems

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.