# 1952 AHSME Problems/Problem 19

## Problem

Angle $B$ of triangle $ABC$ is trisected by $BD$ and $BE$ which meet $AC$ at $D$ and $E$ respectively. Then:

$\textbf{(A) \ }\frac{AD}{EC}=\frac{AE}{DC} \qquad \textbf{(B) \ }\frac{AD}{EC}=\frac{AB}{BC} \qquad \textbf{(C) \ }\frac{AD}{EC}=\frac{BD}{BE} \qquad$

$\textbf{(D) \ }\frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)} \qquad \textbf{(E) \ }\frac{AD}{EC}=\frac{(AE)(BD)}{(DC)(BE)}$

## Solution

$[asy] import olympiad; pair A=(0,0), B=(3,3), C=(6,0), D=(2,0), E=(4,0); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,W); label("B",B,N); label("C",(5.5,0),E); label("D",D,S); label("E",E,S); draw(A--B--C--cycle); draw(B--D); draw(B--E); markscalefactor=0.1; draw(anglemark(A,B,D)); draw(anglemark(D,B,E)); draw(anglemark(E,B,C)); [/asy]$ Using the Angle Bisector Theorem on $\angle ABE$ and $\angle DBC$, we find that $\frac{AD}{DE}=\frac{AB}{BE}$ and $\frac{DE}{EC}=\frac{BD}{BC}$. Rewriting this to better fit our answer choices gives $AD=\frac{(DE)(AB)}{BE}$ and $EC=\frac{(DE)(BC)}{BD}$. Hence, $\frac{AD}{EC}=\frac{(DE)(AB)}{BE}\cdot \frac{BD}{(DE)(BC)}\implies \boxed{\textbf{(D)}\ \frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)}}$.