In the parallelogram ABCD, the point K is the midpoint of BC. It is known that AK = KD.

In the parallelogram ABCD, the point K is the midpoint of BC. It is known that AK = KD. Prove that the given parallelogram is a rectangle.

Consider two triangles ABK and KCD. According to the condition ВK = KС; AK = DK;

Sides AB = CD – opposite sides of the parallelogram. So triangle ABK = triangle KCD on the third basis. Hence <B = <C;

The sum of the angles of a parallelogram adjacent to one side is 180 °.

Means <B = <C = 180 °: 2 = 90 °.

The opposite angles of the parallelogram are equal, that is:

<A = <C = 90 °; <B = <D = 90 °.

Therefore, ABCD is a rectangle