33rd Hellenic Mathematical Olympiad

Let the line $\Delta Z$ meets $\Gamma M,B\Gamma$ at points $K,N$, respectively. Then in the triangle $A\Delta Z$, $B$ is the midpoint of $AZ$ and $BN\parallel A\Delta$. Hence $N$ is the midpoint of $Z\Delta$. Therefore in the triangle $ZE\Delta$, $MN$ connects the midpoints of two sides, and hence:

$$MN//E\Delta (1)$$ Moreover in the triangle $M\Gamma Z$, $\Gamma B,ZK$ are altitudes, and hence $N$ is the ortho-center of the triangle. Hence: $$MN\perp Z\Gamma (2).$$ From (1) and (2) we conclude that $Z\Gamma \perp E\Gamma$. Figure 1