Point $M$ lies on side $B C$ of parallelogram $A B C D$ with an angle of $45^{\circ}$ at vertex $A$, and $\angle A M D=90^{\circ}$, with $B M$ : $M C=2: 3$. Find the ratio of the adjacent sides of the parallelogram.
Apply the Law of Cosines and the Pythagorean Theorem.

## Solution

Let $B M=2 x, C M=3 x, A B=C D=y$. From triangles $A B M$ and $C D M$, using the Law of Cosines, we find that

$$
A M^{2}=4 x^{2}+y^{2}+2 x y \sqrt{2}, D M^{2}=9 x^{2}+y^{2}-3 x y \sqrt{2}
$$

By the Pythagorean Theorem from the right triangle $A M D$, we find that

$$
\begin{gathered}
\left.A M^{2}+M D^{2}=A D^{2}, \text { or } 4 x^{2}+y^{2}+2 x y \sqrt{2}\right)+\left(9 x^{2}+y^{2}-3 x y \sqrt{2}\right)=25 x^{2}, \text { or } \\
12 x^{2}+x y \sqrt{2}-2 y^{2}=0
\end{gathered}
$$

from which $\frac{y}{x}=2 \sqrt{2}$. Therefore,

$$
\frac{A B}{B C}=\frac{y}{5 x}=\frac{2 \sqrt{2}}{5}
$$

## Answer

$2 \sqrt{2}: 5$.