(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:
$$\begin{align*} \angle AED &= 180^\circ - \angle AEB - \angle CED = \\ &= \frac{1}{2}((180^\circ - 2\angle AEB) + (180^\circ - 2\angle CED)) = \\ &= \frac{1}{2}(\angle ABE + \angle DCE) = 90^\circ. \end{align*}$$
So we finished the second part.