[Sum of interior and exterior angles of a polygon]

Vertex $A$ of parallelogram $A B C D$ is connected by segments to the midpoints of sides $B C$ and $C D$. One of these segments turned out to be twice as long as the other. Determine whether angle $B A D$ is acute, right, or obtuse.
Let $N$ be the midpoint of $B C$, $M$ be the midpoint of $C D$, and $A N=2 A M$.

First method. Through point $M$, draw a line parallel to $B C$. It intersects $A B$ at point $K$ and $A N$ at point $P$. Then $K P$ is the midline of triangle $B A N$ (see left diagram) and $A P=1 / 2 A N=A M$. In the isosceles triangle $A P M$, $\angle A M P=\angle A P M$. $\angle P A D=360^{\circ}-(\angle A B M+\angle D M P)-\angle A P M=180^{\circ}-\angle A P M>90^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_a69fc833ac69b1da7555g-33.jpg?height=968&width=1044&top_left_y=12&top_left_x=467)

Second method. Extend segment $A M$ to intersect line $B C$ at point $L$. Triangles $D A M$ and $C L M$ are congruent by side and two adjacent angles (see right diagram). Therefore, $A M=M L$, so $A L=2 A M=A N$. In the isosceles triangle $A N L$, $\angle A N L=\angle A L N90^{\circ}$.

## Answer

Obtuse.