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Iranian Mathematical Olympiad

Geometry Difficulty 6.4 National Olympiad Prove it Iran

Points $D$ and $E$ lie on the side $BC$ of an acute-angled triangle $ABC$ such that $\angle DAB = \angle EAC$ . Let $\omega$ be a circle whose center lies on the circumcircle of the triangle $ABC$ ( $\Gamma$ ) and is tangent to $AD$ at $A$ .

Denote by $A'$ the reflection of point $A$ with respect to $BC$ and the intersection points of $A'E$ and $\omega$ by $K$ and $L$ . Prove that either the lines $BK$ and $CL$ or the lines $BL$ and $CK$ meet on $\Gamma$ .

Solution

Let $F$ be the second intersection point of $\omega$ and $\Gamma$ , and $K'$ the second intersection of $\omega$ and $BF$ . We just need to show that $K'$ lies on $A'E$ . Suppose that $O$ is the center of $\omega$ and $P$ is the intersection point of $AD$ and $\Gamma$ . It is clear that $\angle OAP = 90^\circ$ , therefore $\angle OFP = 90^\circ$ . It yields that $PF$ is tangent to $\omega$ at $F$ and $AP = PF$ . Now notice that
$\angle AFK' = \angle AFB = \angle ACB = \angle ACE, \qquad (1)$
and
$\angle AK'F = 180^\circ - \angle PFA = \angle ABP = \angle AEC. \qquad (2)$
The equations (1) and (2) together imply that the triangles $AK'F$ and $AEC$ are similar. So the triangles $AK'E$ and $AFC$ are similar too. Finally
$\angle AEK' = \angle ACF = \angle APF = 180^\circ - 2\angle AFP = 180^\circ - 2\angle AEB,$
hence the result follows.

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