4. Point E is the midpoint of side AB of parallelogram ABCD. On segment DE, there is a point F such that $\mathrm{AD}=\mathbf{B F}$. Find the measure of angle CFD.
Solution: Extend DE until it intersects line BC at point K (see figure). Since $\mathrm{BK} \| \mathrm{AD}$, then $\angle \mathrm{KBE}=\angle \mathrm{DAE}$. Moreover, $\angle \mathrm{KEB}=\angle \mathrm{DEA}$ and $\mathrm{AE}=\mathrm{BE}$, so triangles $\mathrm{BKE}$ and $\mathrm{ADE}$ are congruent. Therefore, $\mathrm{BK}=\mathrm{AD}=\mathrm{BC}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_92e64e89c4aaddf37f06g-1.jpg?height=369&width=500&top_left_y=1723&top_left_x=367)

Thus, in triangle $\mathrm{CFK}$, the median FB is equal to half the side to which it is drawn, so this triangle is a right triangle with a right angle at F. Therefore, angle CFD is also a right angle.

Answer: $90^{\circ}$.