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Mongolian Mathematical Olympiad Round 3

Geometry Difficulty 5.5 AIME, harder Prove it Mongolia

Let $F$ be a point on the side $AC$ of a triangle $ABC$ . A line through $F$ and parallel to $AB$ intersects with side $BC$ at $D$ . Similarly, a line through $F$ and parallel to $BC$ intersects with side $AB$ at $E$ . Assume that the side $AC$ is tangent to the circumcircle of $EDF$ . If $AB : BC = k$ , then prove that $AF : FC = k^2$ .

Solution

Let $\angle A = \alpha$ , $B = \beta$ , $C = \gamma$ . Then $\angle EFA = \gamma$ , because $EF \parallel BC$ . Since $AC$ is the tangent to the circumcircle of $EDF$ , we have $\angle EDF = \angle EFA = \gamma$ .

Similarly, $\angle DEF = \angle DFC = \alpha$ . It follows that $\triangle EFD \sim \triangle ABC$ .

On the other hand, we have $\triangle AEF \sim \triangle FDC \sim \triangle ABC$ . It follows that $k = \frac{AB}{BC} = \frac{EF}{FD} = \frac{AE}{EF}$ . Hence $FD = EF/k$ . We also have $\triangle EFD \sim \triangle ABC$ . Therefore,
$\frac{AF}{FC} = \frac{AE}{FD} = \frac{AE}{EF/k} = k \frac{AE}{EF} = k^2.$

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