Question 97
We consider the segment $AB$ and the point $\Gamma$ on it such that $AB = 3 \cdot A\Gamma$. We construct the parallelogram $A\Gamma\Delta E$ with $A\Gamma = \Delta E = \Gamma E \gt AE$. We consider point $Z$ on the side $A\Gamma$ such that $\angle AEZ=\angle A\Gamma E=\omega$. Prove that the line perpendicular from $B$ to the line $E\Gamma$ and the line from the point $\Delta$ perpendicular to the line $AB$ meet on a point, say $K$, which lies on the line $EZ$.