8.4. 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 $D E$ to intersect line $B C$ at point $K$ (see Fig. 8.4). Since $B K \| A D$, then $\angle K B E = \angle D A E$. Moreover, $\angle K E B = \angle D E A$ and $A E = B E$, therefore, triangles $B K E$ and $A D E$ are equal. Thus, $B K = A D = B C$.

Therefore, in triangle $CFK$, the median $F B$ is equal to half the side to which it is drawn, so this triangle is a right triangle with a right angle at $F$. Consequently, angle $CFD$ is also a right angle.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ab074edf53f8bb32518g-2.jpg?height=366&width=486&top_left_y=157&top_left_x=1433)

Grading Criteria.

「+」 A correct and well-reasoned solution is provided

「±」 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