Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Estonia

On the side $AB$ of a triangle $ABC$ , let $D$ be a point such that $\angle BDC = \angle ACB$ . Let $K$ be the midpoint of $CD$ and let $E$ be the intersection of lines $BK$ and $AC$ . Given that $\angle BKD = 2\angle BCD$ , find $\angle AEB$ .

Solution

Answer: $90^\circ$ .

Denote $\angle CAB = \alpha$ and $\angle BCA = \gamma$ . Then $\angle BDC = \gamma$ and triangles CBD and ABC will be similar due to having two equal angles (see figure below).

Therefore $\angle BCD = \alpha$ and $\angle BKD = 2\alpha$ . But then $\angle KBC = 2\alpha - \alpha = \alpha = \angle KCB$ , which yields $KB = KC$ . However $KC = KD$ by the choice of $K$ , which yields $\angle KBD = \angle KDB = \gamma$ .

Now triangle $KBD$ gives us $\gamma + \gamma + 2\alpha = 180^\circ$ or $\alpha + \gamma = 90^\circ$ .

Finally, since two angles in triangle $ABE$ are $\alpha$ and $\gamma$ , the third angle $AEB$ must be $180^\circ - (\alpha + \gamma) = 90^\circ$ .

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