37th Hellenic Mathematical Olympiad 2020

a. The triangle $BZ\Gamma$ is right angled at $Z$ and $Z\Delta$ is median. Hence $Z\Delta =\frac{B\Gamma }{2}=B\Delta$. From the isosceles triangle $B\Delta Z$ it follows that: $${\hat{\Delta }}_1=B\hat{\Delta }Z=180^{\circ }-2\hat{B}.(1)$$ Similarly we get $\Delta E=\frac{B\Gamma }{2}=\Gamma \Delta$ and from the isosceles triangle we get: $${\hat{\Delta }}_2=\Gamma \hat{\Delta }E=180^{\circ }-2\hat{\Gamma }.(2)$$

By summing (1) and (2) we find $$\begin{array}{rl}{\hat{\Delta }}_1+{\hat{\Delta }}_2& =(180^{\circ }-2\hat{B})+(180^{\circ }-2\hat{\Gamma })=360^{\circ }-2(\hat{B}+\hat{\Gamma })\\ & =360^{\circ }-2(180^{\circ }-\hat{A})=2\hat{A}\end{array}$$ that is. $$Z\hat{\Delta }E=180^{\circ }-({\hat{\Delta }}_1+{\hat{\Delta }}_2)=180^{\circ }-2\hat{A}.$$ Moreover, since $\Delta ZE$ isosceles ($\Delta Z=\Delta E=\frac{B\Gamma }{2}$), we have: $$\Delta \hat{Z}E=\Delta \hat{E}Z=\frac{180^{\circ }-(180^{\circ }-2\hat{A})}{2}=\hat{A}.$$ figure 1

b. From the triangle $B\Theta Z$ we have: $\hat{B}=B\hat{\Theta }Z+B\hat{Z}\Theta$. (3) and $B\hat{Z}\Theta =A\hat{Z}E$ (4). Moreover, from the cyclic quadrilateral $BZ\Gamma E$ ($BZ\Gamma =B\hat{E}\Gamma =90^{\circ }$) we have: $A\hat{Z}E=\hat{\Gamma }$ (5) Therefore: $B\hat{\Theta }Z=\hat{B}-\hat{\Gamma }$ $$B\hat{\Theta }Z=\hat{B}-\hat{\Gamma }.$$