36th Hellenic Mathematical Olympiad

Question

Let AB quadrilateral inscribed in a circle of center O. The line perpendicular to the side B at its midpoint E meets the line AB at point Z. The circumcircle of the triangle EZ intersects the side AB for a second time at point H and the line at point . The line E meets the line A at point K and the line H at point . Prove that the points A, H, , K are cyclic. FIGURE_0

Accepted Answer

It is enough to prove that: AK = 90^. Since K = AK, it is enough to prove that in the triangle K the two acute angles have sum 90^, i.e. K + K = 90^. We have: aligned K &= E = ZE (inscribed in the same arc) and ZE &= EZB (symmetric with respect to &perpendicular bisector of the side B) aligned Hence we have: K = EZB (1). FIGURE_0 fig. 1 Moreover, from the cyclic quadrilateral AB we have: K = ZBE (2) By summing (1) and (2) we get: K + K = EZB + ZBE = 90^, since the triangle ZBE is right angled at E.