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Geometry Difficulty 4.8 AIME Prove it United States

Problem:

Quadrilateral $ABCD$ satisfies $AB = 8$ , $BC = 5$ , $CD = 17$ , $DA = 10$ . Let $E$ be the intersection of $AC$ and $BD$ . Suppose $BE : ED = 1 : 2$ . Find the area of $ABCD$ .

Solution

Solution:

Since $BE : ED = 1 : 2$ , we have $[ABC] : [ACD] = 1 : 2$ .

Suppose we cut off triangle $ACD$ , reflect it across the perpendicular bisector of $AC$ , and re-attach it as triangle $A' C' D'$ (so $A' = C$ , $C' = A$ ).

Triangles $ABC$ and $C'A'D'$ have vertex $A = C'$ and bases $BC$ and $A'D'$ . Their areas and bases are both in the ratio $1 : 2$ . Thus in fact $BC$ and $A'D'$ are collinear.

Hence the union of $ABC$ and $C'A'D'$ is the $8$ - $15$ - $17$ triangle $ABD'$ , which has area $\frac{1}{2} \cdot 8 \cdot 15 = 60$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.