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Berkeley Math Circle Monthly Contest 2

Geometry Difficulty 4.8 AIME Prove it United States

Problem:
Let $ABC$ be a scalene triangle inscribed in circle $\Gamma$ . The internal bisector of $\angle A$ meets $\overline{BC}$ and circle $\Gamma$ at points $D$ and $E$ . The circle with diameter $\overline{DE}$ meets $\Gamma$ at a second point $F$ . Prove that $\frac{AB}{AC}=\frac{FB}{FC}$ .

Solution

Solution:
Let $K$ be the midpoint of $\operatorname{arc} \widehat{BC}$ containing $A$ . Let $M$ be the midpoint of $\overline{BC}$ .

Since $\angle DME=\angle DFE=90^\circ$ , the points $D, M, E, F$ are concyclic. Moreover, since $\angle EFK=\angle EFD=90^\circ$ , the points $F, D, K$ are collinear. Finally, since $\angle KAD=\angle KMD=90^\circ$ , the points $A, D, M, K$ are concyclic.
Therefore,
$\angle FAD=\angle FAE=\angle FKE=\angle DKM=\angle DAM.$
This implies that $\angle BAF=\angle CAM$ and $\angle CAF=\angle BAM$ . It follows that
$\frac{FB}{FC}=\frac{\sin \angle BAF}{\sin \angle CAF}=\frac{\sin \angle CAM}{\sin \angle BAM}=\frac{AB \cdot \frac{BM}{\sin \angle AMB}}{AC \cdot \frac{CM}{\sin \angle AMC}}=\frac{AB}{AC}$
as required.

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