# Problem №5

Point $E$ is the midpoint of side $A B$ of parallelogram $A B C D$. On segment $D E$, there is a point $F$ such that $A D = B F$. Find the measure of angle $C F D$.
Answer: $90^{\circ}$

Solution

Extend $\mathrm{DE}$ to intersect line $\mathrm{BC}$ at point $\mathrm{K}$ (see figure).

Since $\mathrm{BK} \| \mathrm{AD}$, then $\angle \mathrm{KBE}=\angle \mathrm{DAE}$.

Moreover, $\angle \mathrm{KEB}=\angle \mathrm{DEA}$ and $\mathrm{AE}=\mathrm{BE}$, thus triangles $\mathrm{BKE}$ and $\mathrm{ADE}$ are equal. Therefore, $\mathrm{BK}=\mathrm{AD}=\mathrm{BC}$. Thus, in triangle 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 $\mathrm{F}$. Consequently, angle CFD is also a right angle.

## Grading Criteria.

「+」A correct and justified solution is provided

![](https://cdn.mathpix.com/cropped/2024_05_06_833ed34de7b72168187fg-3.jpg?height=443&width=526&top_left_y=1829&top_left_x=1393)

「土」A generally correct reasoning is provided, containing minor gaps or inaccuracies

「-」Only the answer is provided

「-」The problem is not solved or is solved incorrectly