Parallelogram: diagonals bisect each other

Dependencies: ASA congruence, Parallel ⇒ alternate angles equal (converse), Parallelogram: opposite angles equal / opposite sides equal.

Statement

Let $ABCD$ be a parallelogram (vertices labelled $A\to B\to C\to D$ counterclockwise), so $AB \parallel CD$ and $AD \parallel BC$. Let the two diagonals $AC$ and $BD$ meet at $M$. Then $M$ is simultaneously the midpoint of $AC$ and of $BD$: