2. Points $M, N, P$ are the midpoints of sides $A B, C D$ and $D A$ of the inscribed quadrilateral $A B C D$. It is known that $\angle M P D=150^{\circ}, \angle B C D=140^{\circ}$. Find the angle $\angle P N D$.
- Answer: $110^{\circ}$. Fig. 1. The midline $PN$ is parallel to $AC$, so $\angle PND = \angle ACD = \frac{1}{2} \widehat{AD}$. On the other hand, $\frac{1}{2} \widehat{AB} = \angle ADB = \angle APM = 30^{\circ}$ and $\frac{1}{2} \widehat{BCD} = 40^{\circ}$. Since $\overparen{AD} + \widehat{AB} + \widehat{BCD} = 360^{\circ}$, then $\angle PND = 110^{\circ}$.

(D. Maksimov)

![](https://cdn.mathpix.com/cropped/2024_06_04_fb90edf90df37d91faedg-01.jpg?height=425&width=945&top_left_y=2209&top_left_x=561)

Fig. 1: to problem 2.