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, 2021

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

Let $n$ be the answer to this problem. In acute triangle $A B C$ , point $D$ is located on side $B C$ so that $\angle B A D = \angle D A C$ and point $E$ is located on $A C$ so that $B E \perp A C$ . Segments $B E$ and $A D$ intersect at $X$ such that $\angle B X D = n^{\circ}$ . Given that $\angle X B A = 16^{\circ}$ , find the measure of $\angle B C A$ .

Solution

Solution:

Since $B E \perp A C$ , $\angle B A E = 90^{\circ} - \angle A B E = 74^{\circ}$ . Now, $n^{\circ} = 180 - \angle B X A = \angle E B A + \angle B A D = 16^{\circ} + \frac{74^{\circ}}{2} = 53^{\circ}$ .

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