65th Czech and Slovak Mathematical Olympiad

(i) Let $M$, $N$ be the centers of the legs $BC$, $AD$. We show that the point $N$ lies on the circle with diameter $BC$. A well-known identity yields $$MN=\frac{AB+CD}{2}=\frac{1}{2}BC.$$ It means that the point $N$ has the same distance from the center $M$ of the circle with diameter $BC$ as radius of that circle. So point $N$ lies on that circle.

(ii) With respect to the given condition we can find a point $E$ on the leg $BC$ such that $|BE|=|AB|$ and $|EC|=|CD|$.



The triangles $ABE$, $ECD$ are isosceles and the lines $AB$ and $CD$ are parallel, thus the fact follows: $$\begin{array}{rl}\angle AED& =180^{\circ }-\angle AEB-\angle CED=\\ & =\frac{1}{2}((180^{\circ }-2\angle AEB)+(180^{\circ }-2\angle CED))=\\ & =\frac{1}{2}(\angle ABE+\angle DCE)=90^{\circ }.\end{array}$$ So we finished the second part.