61. In trapezoid $A B C D$, the lateral side $A B$ is perpendicular to $A D$ and $B C$, and $|A B|=\sqrt{|A D| \cdot|B C|}$. Let $E$ be the point of intersection of the non-parallel sides of the trapezoid, $O$ be the point of intersection of the diagonals, and $M$ be the midpoint of $A B$. Find $\widehat{E O M}$.
61. Let $|A D|=a,|B C|=b$. Drop a perpendicular $O K$ from $O$ to $A B$. Now we can find $|B K|=\sqrt{\overline{a b}} \frac{b}{b+a},|B E|=\sqrt{\overline{a b}} \frac{b}{a-b}$, $|M K|=\frac{\sqrt{a b}}{2}-\sqrt{\overline{a b}} \frac{b}{b+a}=\sqrt{\overline{a b}} \frac{a-b}{2(a+b)},|E K|=|B E|+|B K|=$ $=\sqrt{a \bar{b}} \frac{2 a b}{(a-b)(a+b)},|0 K|=\frac{a b}{a+b}$. It is easy to verify that $|O K|^{2}=$ $=|E K| \cdot|M K|$.

Answer: $\widehat{E O M}=90^{\circ}$.