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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.1 AIME, harder Prove it Saudi Arabia

Let $\Gamma$ be a circle with center $O$ and $A E$ be a diameter. Point $D$ lies on segment $O E$ and point $B$ is the midpoint of one of the $\text{arcs A E}$ of $\Gamma$ . Construct point $C$ such that $A B C D$ is a parallelogram. Lines $E B$ and $C D$ meet at $F$ . Line $O F$ meets the minor $\text{arc E B}$ at $I$ . Prove that $E I$ bisects $\angle B E C$ .

Solution

Because $D C$ and $A B$ are parallel, $\angle C D E = \angle B A O = 45^\circ$ . Because $B C$ and $A E$ are parallel, $\angle C B E = \angle O E B = 45^\circ$ . Hence quadrilateral $B C E D$ is cyclic and therefore $\angle B E C = \angle B D C$ .

On the other hand, $\angle D F B = \angle D E F + \angle F D E = 45^\circ + 45^\circ = 90^\circ = \angle A O B$ . We deduce that quadrilateral $B F D O$ is cyclic and therefore $\angle B D C = \angle B O I$ .

But $\angle B O I = 2 \angle B F I$ since both angles intercept the same $\text{arc B I}$ . We conclude that $\angle B E C = 2 \angle B E I$ , which means that $E I$ bisects $\angle B E C$ .

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