Competition problem — Geometry — Olympiad Maths Prep

Geometry Difficulty 6.3 National olympiad Prove it Greece

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

Solution

It is enough to prove that: $A\hat{K}\Lambda = 90^\circ$ .
Since $\Delta\hat{K}\Theta = A\hat{K}\Lambda$ , it is enough to prove that in the triangle $\Delta\Theta K$ the two acute angles have sum $90^\circ$ , i.e. $\Delta\Theta K + \Theta\hat{\Delta}K = 90^\circ$ . We have:
$\begin{aligned} \Delta\Theta K &= \Gamma\Theta E = \Gamma\hat{Z}E \quad (\text{inscribed in the same arc}) \text{ and} \\ \Gamma\hat{Z}E &= E\hat{Z}B \quad (\text{symmetric with respect to} \\ &\text{perpendicular bisector of the side } B\Gamma) \end{aligned}$
Hence we have: $\Delta\Theta K = E\hat{Z}B$ (1).

fig. 1

Moreover, from the cyclic quadrilateral $AB\Gamma\Delta$ we have: $\Theta\hat{\Delta}K = Z\hat{B}E$ (2)
By summing (1) and (2) we get: $\Delta\Theta K + \Theta\hat{\Delta}K = E\hat{Z}B + Z\hat{B}E = 90^\circ$ , since the triangle $ZBE$ is right angled at $E$ .

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