Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Find the answer

In triangle $A B C, A C=3 A B$ . Let $A D$ bisect angle $A$ with $D$ lying on $B C$ , and let $E$ be the foot of the perpendicular from $C$ to $A D$ . Find $[A B D] /[C D E]$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

By the Angle Bisector Theorem, $D C / D B=A C / A B=3$ . We will show that $A D=$ $D E$ . Let $C E$ intersect $A B$ at $F$ . Then since $A E$ bisects angle $A, A F=A C=3 A B$ , and $E F=E C$ . Let $G$ be the midpoint of $B F$ . Then $B G=G F$ , so $G E \| B C$ . But then since $B$ is the midpoint of $A G, D$ must be the midpoint of $A E$ , as desired. Then $[A B D] /[C D E]=(A D \cdot B D) /(E D \cdot C D)=1 / 3$ .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.