Hellenic Mathematical Olympiad

From the relationship of a central angle and an inscribed angle that go on the same arc $\Delta K$ of the circle $\gamma$, we have: $$\angle A\Delta K=2\cdot \angle E\Delta K(1)$$ Also we have the equality of inscribed angles $$\angle E\Delta K=\angle AB\Delta (2)$$ From relations (1) and (2) it follows that $\angle A\Delta K=2\cdot \angle AB\Delta$, that is, $AB$ is the bisector of the angle $\angle A\Delta K$.

Figure 1

Since $A\Delta =AK$, (radii of the circle $\gamma$), the triangle $\Delta AK$ is isosceles and hence $AB\perp \Delta K$, that is $AB\perp \Gamma \Delta$. Similarly we get that $A\Gamma \perp BE$. Hence $K$ is the orthocenter of the triangle $AB\Gamma$, and therefore $AK\perp B\Gamma$.