Prove that $PBRS$ is a parallelogram.
(Note: $P$ and $Q$ are respectively the middles of the sides $AB$ and $CD$)
Now the corrections give the following method:
- $PBQD$ is a parallelogram
- $BR \parallel PS$
- $\triangle BPR \cong \triangle SRP$ (SAS) $\Longrightarrow$ $\angle BPR=\angle SRP$ (alternate interior angles) $\Longrightarrow BP\parallel RS$
- In quadrilateral $PBRS$, opposite sides are equal in length, so $PBRS$ is a parallelogram.
Since I find the first line a bit vague, is there another way to approach that part of the problem that also uses the congruent triangles?