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Junior Turkish Mathematical Olympiad

Geometry Difficulty 4.0 AIME Prove it Turkey

A circle that passes through the vertex $A$ of a rectangle $ABCD$ intersects the side $AB$ at a second point $E$ different from $B$ . A line passing through $B$ is tangent to this circle at a point $T$ , and the circle with center $B$ and passing through $T$ intersects the side $BC$ at the point $F$ . Show that if $\angle CDF = \angle BFE$ , then $\angle EDF = \angle CDF$ .

Solution

Let $G$ be the point of intersection of the lines $EF$ and $DC$ . Since $\angle CFG = \angle BFE = \angle CDF$ , the lines $DF$ and $EG$ are perpendicular.

On the other hand, $BA \cdot BE = BT^2 = BF^2$ implies that the triangles $BAF$ and $BFE$ are similar and $\angle BAF = \angle BFE = \angle CDF$ . Therefore $BF = CF$ , $EF = GF$ and $\angle EDF = \angle GDF = \angle CDF$ .

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