# Proof with parallelogram inside a parallelogram

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?

Another proof: $$\triangle ADP\cong \triangle CBQ$$ because of SAS, because $|AD|=|BC|$, $|AP|=\frac 12|AB|=\frac 12 |CD|=|CQ|$ and $\angle DAP=\angle DAB=\angle BCD=\angle BCQ$. This implies $\angle ADP=\angle CBQ$. Now $DP$ and $BQ$ are parallel because $AD$ and $BC$ are parallel. Now, use the second part of the proof you gave.