2. The sides $\overline{B C}$, $\overline{C D}$, and $\overline{D A}$ of the isosceles trapezoid $A B C D$ are congruent to each other. Squares $A D E F$ and $C B G H$ are drawn on its legs from the outside. If the obtuse angle of the trapezoid is seven times larger than its acute angle, determine the measure of the angle formed by the lines $E F$ and $G H$.
## First Solution.

If $\alpha$ is the measure of the acute angle and $\beta$ is the measure of the obtuse angle of the trapezoid $ABCD$, according to the problem, we have $\beta=7\alpha$. The sum of the interior angles of a quadrilateral is $360^{\circ}$, and in an isosceles trapezoid, the acute angles are congruent to each other and the obtuse angles are congruent to each other, so we have:

$$
2\alpha + 2\beta = 360^{\circ} \quad \alpha + \beta = 180^{\circ} \quad \alpha + 7\alpha = 180^{\circ} \quad 8\alpha = 180^{\circ}
$$

from which we get $\alpha = 22.5^{\circ}$ and $\beta = 157.5^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_ff246218007c389a8ad8g-3.jpg?height=408&width=1556&top_left_y=767&top_left_x=250)

Let the lines $EF$ and $GH$ intersect at point $S$. Let the line on which the base $\overline{AB}$ lies intersect the line $EF$ at point $K$ and the line $GH$ at point $L$. We want to determine the angle $|\angle KSL|$.

In $\triangle AFK$, we have $|\angle KFA| = 90^{\circ}$ and $|\angle FAK| = 180^{\circ} - |\angle DAF| - |\angle BAD| = 180^{\circ} - 90^{\circ} - 22.5^{\circ} = 67.5^{\circ}$.

Therefore, we conclude $|\angle AKF| = 180^{\circ} - |\angle KFA| - |\angle FAK| = 22.5^{\circ}$.

Similarly, we calculate $|\angle GBL| = 22.5^{\circ}$. Thus, $|\angle LKS| = |\angle SLK| = 22.5^{\circ}$.

It follows that $|\angle KSL| = 180^{\circ} - 22.5^{\circ} - 22.5^{\circ} = 135^{\circ}$. The desired angle between the lines is $45^{\circ}$.

Note. Instead of calculating the angles in triangle $AFK$, we can observe that the lines $AD$ and $KS$ are parallel (since the line $KS$ contains the side $\overline{EF}$ which is opposite to the side $\overline{AD}$), and conclude that $|\angle AKF| = |\angle BAD| = 22.5^{\circ}$ due to the equality of the angles on the transversal.