Solution

Since $\overline{A E}$ is the diagonal of the square $A B E F$, then $\angle A E F=45^{\circ}$. Extend the segment $\overline{A E}$ to form the right triangle $M P D$ as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-38.jpg?height=740&width=959&top_left_y=460&top_left_x=435)

Notice that $\measuredangle P E D=\measuredangle A E F=45^{\circ}$, and therefore, the right triangle $E P D$ is isosceles. The hypotenuse of the triangle $E P D$ measures 4 (which is the side of the square $B C D E$). Thus, the legs measure $|E P|=|P D|=2 \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-38.jpg?height=745&width=959&top_left_y=1428&top_left_x=435)

Finally, $\angle M D P=(15+45)^{\circ}$, and therefore, in the right triangle $M P D$ the internal angles are $30^{\circ}$ and $60^{\circ}$, and the following relationship holds:

$$
|M D|=2|P D|
$$

Thus, we have proven that $|M D|=2(2 \sqrt{2})=4 \sqrt{2}$.
In the following figure, $A B E F$ and $E B C D$ are squares.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-11.jpg?height=465&width=832&top_left_y=460&top_left_x=689)

If $\measuredangle M D E=15^{\circ}$ and $|A F|=4$, calculate $|M D|$.

Note: $|M D|$ is the length of the segment $M D$.

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