4. Let $E$ be the midpoint of side $CD$ of square $ABCD$. Point $M$ inside the square is such that $\measuredangle MAB = \measuredangle MBC = \measuredangle BME = x$. Determine $x$?
Solution. Let's denote $\measuredangle A B M$ as $y$. Since $\measuredangle A B C=x+y=90^{\circ}$, from triangle $A B M$ we get that $\measuredangle A M B=90^{\circ}$. Let $F$ be the midpoint of side $A B$. Then from the right triangle $A B M$ we get
$\overline{A F}=\overline{F B}=\overline{M F}$. Furthermore, $\measuredangle F M B=\measuredangle F B M=y$ from which it follows that $\measuredangle E M F=x+y=90^{\circ}$. Triangle $M F E$ is a right triangle where $E F=2 M F$ from which it follows that $\measuredangle M F E=60^{\circ}$ and

$$
\measuredangle M B F=\frac{1}{2} \measuredangle M F A=\frac{1}{2}\left(90^{\circ}-\measuredangle M F E\right)=15^{\circ}
$$

Finally, $x=75^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_d4a9105d103058c5b923g-08.jpg?height=389&width=334&top_left_y=260&top_left_x=1112)