7. Let $A B C D$ be a rectangle and let $M, N$ be points inside, respectively, the sides $A B$ and $B C$. Suppose that $M C = C D, M D = M N$ and that the points $C, D, M, N$ lie on the same circle.

What is the ratio $A B / B C$?
(A) $\frac{1+\sqrt{3}}{2}$
(B) $\sqrt{2}$
(C) $\frac{3}{2}$
(D) $\sqrt{3}$
(E) 2
7. The answer is (B). Since the quadrilateral $C D M N$ is circumscribed, we know that the angle $\angle C D M$ is equal to the supplement of the angle $\angle M N C$, that is, to the angle $\angle M N B$. On the other hand, the angles $\angle C D M$ and $\angle A M D$ are equal, as they are alternate interior angles formed by the parallel lines $A B$ and $C D$ with the transversal $D M$. We therefore deduce that $\angle A M D=\angle M N B$.

The right triangles $A D M$ and $B M N$ thus have an equal acute angle and, by hypothesis, equal hypotenuses $(D M=M N)$ and are therefore congruent. In particular, it follows that $A D=M B$, and, since $A D=B C$, the triangle $B C M$ is isosceles with respect to the base $C M$ (in addition to being right-angled at $B$), so the ratio between its hypotenuse and one of the legs is $\sqrt{2}$. Therefore, $A B / B C=C D / B C=C M / B C=\sqrt{2}$.