SELECTION EXAMINATION

Since $\Delta ,E,Z$ are the midpoints of the sides $B\Gamma ,A\Gamma ,AB$, respectively, the quadrilaterals $AE\Delta Z,ZE\Delta B$ and $ZE\Gamma \Delta$ are parallelograms. From $\Delta M\parallel \Gamma \Sigma$ and $MN\parallel K\Lambda$ we conclude that: $\hat{M}={\hat{\Sigma }}_1$ and ${\hat{\Sigma }}_1=\hat{K}$. From the cyclic quadrilateral $B\Lambda K\Gamma$ ($B\Lambda \Gamma =B\hat{K}\Gamma =90^{\circ }$) we get: $\hat{K}=\hat{B}$.

From the last three equalities of angles we arrive at the equality $\hat{M}=\hat{B}$, and from this we conclude that the points $M,B,\Delta ,T$, are cocyclic.

In a similar fashion, since $\Delta N//BT$ and $MN//K\Lambda$ we have the equalities $\hat{N}={\hat{T}}_1$ and ${\hat{T}}_1=\hat{\Lambda }$. Also, from the cyclic quadrilateral $B\Lambda K\Gamma$ (since $B\hat{\Lambda }\Gamma =B\hat{K}\Gamma =90^{\circ }$) we have that: $\hat{\Lambda }=\hat{\Gamma }$. From the last three equalities of angles we have $\hat{N}=\hat{\Gamma }$, and from this we conclude that the points $N,\Gamma ,\Delta ,\Sigma$, are cocyclic.

Since the points $B,\Delta ,P,T,M$ belong to the circle $(c_1)$, from the cyclic quadrilateral $MTP\Delta$ we have: ${\hat{T}}_2+\hat{\Delta }=180^{\circ }$. Since the points $\Gamma ,\Delta ,\Pi ,\Sigma ,N$ belong to the circle $(c_2)$, from the cyclic quadrilateral $\Delta \Pi \Sigma N$ we have: ${\hat{\Sigma }}_2+\hat{\Delta }=180^{\circ }$. From the last two equalities we get: $${\hat{T}}_2={\hat{\Sigma }}_2=180^{\circ }-\hat{\Delta }=180^{\circ }-\hat{A}.$$

The quadrilateral $MTP\Delta$ is inscribed to the circle $(c_1)$, and hence: $$\widehat{TPE}=\hat{P}\Delta =180^{\circ }-\hat{M}\Delta =180^{\circ }-\hat{B}\Gamma =180^{\circ }-\hat{B}.$$

The angle ${\hat{\Sigma }}_2=180^{\circ }-\hat{A}$ is an external angle of the triangle $\Sigma \Pi M$, and so: $$\Sigma \hat{\Pi }Z=\Sigma \hat{\Pi }M={\hat{\Sigma }}_2-\hat{M}=180^{\circ }-\hat{A}-\hat{B}=\hat{\Gamma }=\Sigma \hat{E}Z.$$ The quadrilateral $\Sigma \Pi \Gamma \Pi$ is inscribed into the circle $(c)$ and since ${\hat{T}}_2={\hat{\Sigma }}_2$ it is isosceles trapezium and so $\Sigma T\parallel \Pi P$.