Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it United States

Problem:

In triangle $ABC$ , $AC = 3 AB$ . Let $AD$ bisect angle $A$ with $D$ lying on $BC$ , and let $E$ be the foot of the perpendicular from $C$ to $AD$ . Find $[ABD]/[CDE]$ . (Here, $[XYZ]$ denotes the area of triangle $XYZ$ .)

Solution

Solution:

$1/3$

By the Angle Bisector Theorem, $DC/DB = AC/AB = 3$ . We will show that $AD = DE$ . Let $CE$ intersect $AB$ at $F$ . Then since $AE$ bisects angle $A$ , $AF = AC = 3 AB$ , and $EF = EC$ . Let $G$ be the midpoint of $BF$ . Then $BG = GF$ , so $GE \parallel BC$ . But then since $B$ is the midpoint of $AG$ , $D$ must be the midpoint of $AE$ , as desired. Then $[ABD]/[CDE] = (AD \cdot BD)/(ED \cdot CD) = 1/3$ .

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