5*. In rhombus $A B C D$ the measure of angle $B$ is $40^{\circ}, E-$ is the midpoint of $B C, F$ - is the foot of the perpendicular dropped from $A$ to $D E$. Find the measure of angle $D F C$.
5. Let lines $D E$ and $A B$ intersect at point $G$ (Fig. 83). Then triangles $D E C$ and $B E G$ are congruent by the second criterion. Therefore, $B G=C D=B A$. Hence, points $A, G$, and $C$ lie on a circle with center at point $B$, and $A G$ is the diameter. Since $\angle A F G=90^{\circ}$, point $F$ lies on the same circle by the theorem about angles subtended by a diameter (see fact 14).

![](https://cdn.mathpix.com/cropped/2024_05_21_6455f48598646f41a890g-193.jpg?height=228&width=490&top_left_y=306&top_left_x=754)

Fig. 83

By the inscribed angle theorem, we have $\angle G F C=\frac{1}{2} \angle G B C=\frac{1}{2}\left(180^{\circ}-40^{\circ}\right)=70^{\circ}$. Therefore, $\angle D F C=180^{\circ}-$ $-\angle G F C=110^{\circ}$.

Alternative solution. We can solve the problem without using circles. It is known that the median of a right triangle drawn to the hypotenuse is equal to half the hypotenuse (see fact 14). Applying this statement to triangle $A F G$, we get $B F=B A=B G$.

Triangles $C B F$ and $F B A$ are isosceles, so the sum of angles $B C F$ and $B A F$ equals angle $C F A$.

The sum of the angles in quadrilateral $A B C F$ is $360^{\circ}$. Therefore,

$$
\angle C F A+\angle C F A+40^{\circ}=360^{\circ},
$$

from which $\angle C F A=160^{\circ}$. Hence, $\angle C F D=360^{\circ}-\angle A F C-$ $-\angle A F D=360^{\circ}-160^{\circ}-90^{\circ}=110^{\circ}$.