Since angle $D K F$ is an exterior angle for triangle $D K E$, we have $\angle D E K = \angle D K F - \angle K D E = 70^{\circ}$ (see the figure). Therefore, triangle $D K E$ is isosceles:

$D K = E K = F K$.

Thus, the median $D K$ of triangle $D E F$ is equal to half the side $E F$ to which it is drawn, so $\angle E D F = 90^{\circ}$. Consequently, $\angle D F E = 180^{\circ} - (\angle D E F + \angle E D F) = 20^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_34fd2032e75750f56a92g-08.jpg?height=329&width=736&top_left_y=1136&top_left_x=681)

## Answer

$70^{\circ} ; 90^{\circ}$ and $20^{\circ}$.
In triangle $D E F$, the median $D K$ is drawn. Find the angles of the triangle if $\angle K D E=70^{\circ}, \angle D K F=140^{\circ}$.