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Estonia — Open Contests

Geometry Difficulty 4.6 AIME Prove it Estonia

A circle $c$ with center $A$ passes through the vertices $B$ and $E$ of a regular pentagon $ABCDE$ . The line $BC$ intersects the circle $c$ the second time at point $F$ . Prove that lines $DE$ and $EF$ are perpendicular.

Solutions — 2

Solution 1

The internal angles of a regular pentagon have size $108^\circ$ . Thus $\angle EAB = 108^\circ$ (Fig. 2), whence $\angle EFC = \angle EFB = \frac{\angle EAB}{2} = 54^\circ$ . As $\angle CDE = 108^\circ$ and $\angle FCD = \angle BCD = 108^\circ$ , from the quadrilateral $CDEF$ we obtain $\angle DEF = 360^\circ - \angle FCD - \angle CDE - \angle EFC = 90^\circ$ .

Fig. 2

Solution 2

The internal angles of a regular pentagon have size $108^\circ$ . Thus $\angle ABC = 108^\circ$ , whence $\angle ABF = 180^\circ - \angle ABC = 72^\circ$ . As $AB = AF$ , from the triangle $ABF$ we obtain $\angle BAF = 180^\circ - 2 \cdot 72^\circ = 36^\circ$ . Let $G$ be the second intersection point of line $DE$ with circle $c$ (Fig. 3); by symmetry, $\angle EAG = \angle BAF = 36^\circ$ . Since $\angle EAB = 108^\circ$ , we have $\angle FAG = \angle BAF + \angle EAB + \angle EAG = 180^\circ$ , i.e., $FG$ is a diameter of $c$ . Hence $\angle FEG = 90^\circ$ .

Fig. 3

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