Problem 10.3. On the side $AD$ of rectangle $ABCD$, a point $E$ is marked. On the segment $EC$, there is a point $M$ such that $AB = BM, AE = EM$. Find the length of side $BC$, given that $ED = 16, CD = 12$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d55ea3e8288d58a3f771g-37.jpg?height=367&width=497&top_left_y=93&top_left_x=476)
Answer: 20.

Solution. Note that triangles $A B E$ and $M B E$ are equal to each other by three sides. Then $\angle B M E=\angle B A E=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d55ea3e8288d58a3f771g-37.jpg?height=361&width=495&top_left_y=659&top_left_x=479)

Fig. 6: to the solution of problem 10.3

From the parallelism of $A D$ and $B C$, it follows that $\angle B C M=\angle C E D$ (Fig. 6). Therefore, right triangles $B C M$ and $C E D$ are equal by an acute angle and the leg $B M=A B=C D$. Using the Pythagorean theorem for triangle $C D E$, we conclude

$$
B C=C E=\sqrt{C D^{2}+E D^{2}}=\sqrt{12^{2}+16^{2}}=20
$$