2. Point $\mathbf{E}$ 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 $\mathrm{DE}$ to intersect line $\mathrm{BC}$ at point $\mathrm{K}$ (see the figure). Since $\mathrm{BK} \| \mathrm{AD}$, we have $\angle \mathrm{KBE} = \angle \mathrm{DAE}$. Additionally, $\angle \mathrm{KEB} = \angle \mathrm{DEA}$ 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_9e95ff2d962ae0521da7g-1.jpg?height=366&width=502&top_left_y=1279&top_left_x=366)

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

Answer: $90^{\circ}$.