Selection Examination

a. The angles $\widehat{AMB}$ and $\widehat{AN\Gamma }$ are right since they see the diameters $AB$ and $A\Gamma$. Therefore the quadrilateral $AMHN$ is cyclic, which is the desired result.



b. ${\hat{T}}_1={\hat{I}}_1(1)$.

The line $E\Delta$ connects the midpoints of $AB$ and $A\Gamma$, so $MN$ is parallel to $B\Gamma$. Therefore:

$${\hat{r}}_1={\hat{N}}_1(2).$$

From the quadrilateral $AMHN$ we have: $${\hat{H}}_1={\hat{N}}_2(3).$$ From the relations (1), (2), (3) we have: $${\hat{T}}_1+{\hat{H}}_1={\hat{N}}_1+{\hat{N}}_2=A\hat{N}H=90^{\circ }.$$



From the quadrilateral $AB\Sigma \Gamma$ we have: $$A\hat{B}\Gamma =A\hat{\Sigma }\Gamma =\hat{B}\text{and from }E\Delta \parallel B\Gamma \text{ we have: }{\hat{E}}_1=\hat{B}.$$ From the equality $A\hat{\Sigma }\Gamma =A\hat{\Sigma }N$ in combination with the previous ones we get that the quadrilateral $AE\Sigma N$ is cyclic.

The quadrilateral $AZ\Sigma N$ is also cyclic, since $A\hat{N}\Sigma =A\hat{Z}\Sigma =90^{\circ }$. To this end, we have that ${\hat{Z}}_1={\hat{N}}_2={\hat{H}}_1$ and so $BH\parallel EZ$. Since $E$ is the midpoint of $AB$, we conclude that $Z$ is the midpoint of $AH$.