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, $\frac{AD}{EC}=\frac{(DE)(AB)}{BE}\cdot \frac{BD}{(DE)(BC)}\implies \boxed{\textbf{(D)}\ \frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)}}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1952 AHSME Problems/Problem 19

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