4. Let square $A B C D$. An equilateral triangle $\triangle A B E$ is constructed over side $\overline{A B}$, such that point $E$ is inside the square. Line $D E$ intersects side $\overline{B C}$ at point $F$. What is the measure of angle $\angle B E F$?
4. Let $ABCD$ be a square. An equilateral triangle $ABE$ is constructed over the side $\overline{AB}$, such that point $E$ is inside the square. The line $DE$ intersects the side $\overline{BC}$ at point $F$. What is the measure of the angle $\Varangle BEF$?

## Solution.

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_1ae5e90785d026149547g-25.jpg?height=553&width=808&top_left_y=603&top_left_x=630)

Since $|AD| = |AB| = |AE|$, we conclude that $\triangle AED$ is isosceles.

This means that $|\Varangle DEA| = |\Varangle ADE|$

1 POINT

Furthermore,

$|\Varangle EAD| = |\Varangle BAD| - |\Varangle BAE| = 90^{\circ} - 60^{\circ} = 30^{\circ}$

Then from $\triangle AED$ it follows:

$|\measuredangle DEA| = (180^{\circ} - 30^{\circ}) : 2 = 150^{\circ} : 2 = 75^{\circ}$

Points D, E, and F are on the same line, so:

$|\Varangle DEA| + |\Varangle AEB| + |\Varangle BEF| = 180^{\circ}$

$75^{\circ} + 60^{\circ} + |\Varangle BEF| = 180^{\circ}$

$|\Varangle BEF| = 180^{\circ} - 135^{\circ} = 45^{\circ}$

The measure of the angle $\Varangle BEF$ is $45^{\circ}$.