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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.6 AIME Prove it United States

Problem:

In trapezoid $ABCD$ shown, $AD$ is parallel to $BC$ , and $AB = 6$ , $BC = 7$ , $CD = 8$ , $AD = 17$ . If sides $AB$ and $CD$ are extended to meet at $E$ , find the resulting angle at $E$ (in degrees).

Solution

Solution:

Choose point $F$ on $AD$ so that $BCDF$ is a parallelogram. Then $BF = CD = 8$ , and $AF = AD - DF = AD - BC = 10$ , so $\triangle ABF$ is a $6$ - $8$ - $10$ right triangle. The required angle is equal to $\angle ABF = 90^{\circ}$ .

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