7. In rectangle $A B C D$, point $E$ is located on diagonal $A C$ such that $B C=E C$, point $M$ is on side $B C$ such that $E M=M C$. Find the length of segment $M C$, if $B M=5, A E=2$.
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_de39ee82fbf77ee5c736g-6.jpg?height=266&width=477&top_left_y=898&top_left_x=915)

Draw $A F$ parallel to $B E$ (point $F$ lies on line $B C$), then $\angle C B E=\angle C F A$, $\angle C E B=\angle C A F$. Considering that $B C=C E$, we get that triangle $FCA$ is isosceles, hence $F C=A C$ and $F B=A E$. Triangles $F B A$ and $A E F$ are congruent, as $F B=A E, \angle A F B=$ $\angle F A E, A F$ is common. We obtain that $\angle F B A=\angle A E F=90^{\circ}$, from which $\angle F E C=90^{\circ}$. Triangle $F C E$ is right-angled and $M C=M E$, so $\mathrm{FM}=\mathrm{MC}$ and $F M=F B+B M=A E+B M=M C=7$.

Answer: 7.