The figure shows a rectangle $A B C D$ with side $\overline{A D}=4$ and where point $M$ is the midpoint of segment $A B$.

![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-19.jpg?height=508&width=708&top_left_y=500&top_left_x=777)

Knowing that segments $A C$ and $D M$ are orthogonal, calculate the area of rectangle $A B C D$.

#
Solution

Let's call $P$ the intersection point of segments $A C$ and $D M$. Since the sum of the internal angles of triangle $A P M$ is $180^{\circ}$, then

$$
\measuredangle M A P=90^{\circ}-\measuredangle A M P
$$

But $\measuredangle M A P+\measuredangle P A D=90^{\circ}$. Therefore, we conclude that $\measuredangle P A D=\measuredangle A M P$. We have thus shown that the right triangles $M A D$ and $A D C$ have the same internal angles. They are therefore similar.

![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-48.jpg?height=391&width=549&top_left_y=1723&top_left_x=662)

Applying the similarity between these triangles, we have:

$$
\frac{\overline{A D}}{\overline{D C}}=\frac{\overline{M A}}{\overline{A D}}
$$

We know that $\overline{A D}=4$ and that $\overline{D C}=2 \times \overline{M A}$. Using these facts in (.21) we obtain $\overline{M A}=2 \sqrt{2}$ and therefore $\overline{D C}=4 \sqrt{2}$. Finally, the area of rectangle $A B C D$ is $\overline{A D} \times \overline{D C}=16 \sqrt{2}$.