2. Let there be a rectangle $A B C D$. The perpendicular bisector of the diagonal $\overline{A C}$ intersects the side $\overline{A B}$ at point $E$, and the side $\overline{C D}$ at point $F$ such that the triangle $E B C$ is isosceles. Determine the measure of the angle $\measuredangle D F E$.
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_c165b3842bf96502feaeg-09.jpg?height=306&width=554&top_left_y=1149&top_left_x=291)

Since triangle $E B C$ is an isosceles right triangle, it follows that $|\measuredangle E C B|=|\measuredangle B E C|=45^{\circ}$.

Since $A B C D$ is a rectangle, it follows that $|\measuredangle F C E|=90^{\circ}-|\measuredangle E C B|=45^{\circ}$.

Since point $E$ lies on the bisector $E F$ of the diagonal $\overline{A C}$, it follows that $|E A|=|E C|$, which means that triangle $A E C$ is isosceles.

Thus, $|\measuredangle A C E|=|\measuredangle E A C|$.

Furthermore, $|\measuredangle E A C|=|\measuredangle F C A|$ because they are acute angles with parallel sides.

Therefore, $|\measuredangle A C E|=|\measuredangle F C A|=\frac{|\measuredangle F C E|}{2}=\frac{45^{\circ}}{2}=22.5^{\circ}$.

Since $\measuredangle D F E$ is an exterior angle of triangle $C F S$, it follows that

$|\measuredangle D F E|=|\measuredangle C S F|+|\measuredangle F C A|=90^{\circ}+22.5^{\circ}=112.5^{\circ}$