, 2020

If $A\Gamma E = \omega$, from the right angled triangle $\Gamma\Theta B$ we have:
$$\Gamma \hat{B} \Theta = 90^{\circ} - B \hat{\Gamma} \Theta = 90^{\circ} - \omega, \qquad (1)$$
since $B\Gamma\Theta = A\Gamma E = \omega$, and moreover $B\hat{E}K = 90^{\circ}$, and so
$$\Gamma \hat{B} \Theta + B \hat{H} K = 90^{\circ} - \omega + 90^{\circ} = 180^{\circ} - \omega < 180^{\circ},$$
Which gives that the lines $B\Theta$ and $\Delta H$ intersect at a point, say $K$. Therefore, it is enough to prove that the points $Z$, $E$ and $K$ are collinear.
The quadrilateral $E\Delta\Gamma Z$ is isosceles trapezium, because $E\Delta \parallel Z\Gamma$ and $ZE = AE = \Gamma\Delta$. Indeed, the triangles $AEZ$ and $A\Gamma E$ have the angle $A$ in common and $A\hat{E}Z = A\Gamma E = \omega$, therefore they have equal angles. Therefore the triangle $AZE$ is isosceles with $A\hat{E}Z = E\hat{A}Z$. It follows that: $AE = ZE$.

figure 2
From the isosceles trapezium $E\Delta\Gamma Z$ we have that:
$$E \hat{Z} \Gamma = Z \hat{\Gamma} \Delta = 180^{\circ} - \Gamma \hat{A} E = 180^{\circ} - \left( \frac{180^{\circ} - \omega}{2} \right) = 90^{\circ} + \frac{\omega}{2}. \qquad (2)$$
From the orthogonal triangle $\Delta H Z$ we have
$$Z\hat{\Delta}H = 90^{\circ} - \Delta\hat{Z}H = 90^{\circ} - \omega. \qquad (3)$$
From (1) and (3) it follows that the quadrilateral $B\Delta ZK$ is cyclic.
If $\Lambda$ is the midpoint of the segment $\Gamma B$ we observe that the quadrilateral $E\Delta\Lambda\Gamma$ has $E\Delta \parallel \Gamma\Lambda$, $\Gamma E = \Delta E = A\Gamma = \Gamma\Lambda$, since $A\Gamma = \frac{1}{3}AB$ and $\Lambda$ is the midpoint of $\Gamma B$. Hence $E\Delta\Lambda\Gamma$ is rhombus. Hence, we have $\Lambda\Gamma\Delta = \omega$ and moreover $\Delta\Lambda = \Lambda\Gamma = \Lambda B = \frac{\Gamma B}{2}$. It means that the triangle $\Gamma\Delta B$ is orthogonal at $\Delta$ and from the isosceles triangle $\Lambda\Delta B$ we have : $\Delta\hat{B}\Lambda = \frac{\omega}{2}$.
From the cyclic quadrilateral $B\Delta ZK$ we have:
$$Z\hat{K}\Delta = \Delta\hat{B}Z = \Delta\hat{B}\Lambda = \frac{\omega}{2}, \qquad (4)$$
οπότε από το ορθογώνιο τρίγωνο $HZK$ και τη σχέση (4) έπεται ότι:
$$\Gamma\hat{Z}K = H\hat{Z}K = 90^{\circ} - \frac{\omega}{2} \qquad (5)$$
From (2) and (5) it follows that:
$$E\hat{Z}\Gamma + \Gamma\hat{Z}K = 90^{\circ} + \frac{\omega}{2} + 90^{\circ} - \frac{\omega}{2} = 180^{\circ},$$
and hence the points $A$, $E$ and $K$ are collinear.