In parallelogram ABCD BC: AB = 1: 2. The middle M of side AB is connected by segments

In parallelogram ABCD BC: AB = 1: 2. The middle M of side AB is connected by segments with vertices C and D. Prove that the angle CMD is equal to 90 degrees.

Since | AB | = 2 | ВС | and M is the middle of AB, then
| AM | = | MB | = | AB | / 2 = | BC | = | AD |
Accordingly, in the MBC triangle:
| MB | = | ВС |,
and MBC is an isosceles triangle. Therefore:
∠ВМС = 90 ° – ∠В / 2
Similar:
| AM | = | AD |,
АМD is an isosceles triangle, and:
∠АМD = 90 ° – ∠А / 2
As:
∠АМD + ∠СМD + ∠ВМС = 180 °,
then
∠СМD = 180 ° – ∠АМD – ∠ВМС = 180 ° – (90 ° -∠А / 2) – (90 ° -∠В / 2) = (∠А + ∠В) / 2
ABCD – parallelogram, and:
∠А + ∠В = 180 °
As a result, we get:
∠СМD = (∠А + ∠В) / 2 = 180 ° / 2 = 90 °,
Q.E.D.