2023/24 AHSMC Part 1 Problem 14

Problem 14 of 16

Let $ABCD$ be a convex quadrilateral whose diagonals $AC$ and $BD$ intersect at point $E$ . If $BC = 6$ , $DA = 8$ and the triangles $\triangle AEB$ and $\triangle CED$ have the same area, what is the ratio of the area of $\triangle BEC$ over the area of $\triangle DEA$ ?

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Solution

Since $\mathrm{Area}(\triangle AEB) = \mathrm{Area}(\triangle CED)$ , we have $\mathrm{Area}(\triangle ABD) = \mathrm{Area}(\triangle AEB) + \mathrm{Area}(\triangle DEA) = \mathrm{Area}(\triangle CED) + \mathrm{Area}(\triangle DEA) = \mathrm{Area}(\triangle CDA).$ So the distances from $B$ and $C$ to the line $DA$ are equal, hence $BC \parallel DA$ and $ABCD$ is a trapezoid. Then $\triangle BEC \sim \triangle DEA$ and $\frac{\mathrm{Area}(\triangle BEC)}{\mathrm{Area}(\triangle DEA)} = \left(\frac{BC}{DA}\right)^2 = \frac{9}{16}.$ The answer is (E).

University of Alberta

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