similar triangle problem in parallelogram with vertical lines

Question

Can anyone help me with this task? I have no idea how to start.
From the top $B$ of a parallelogram $ABCD$ lowered the vertical $BP$ and $BQ$ on the directions of $AD$ and $CD$ . From the top $D$ parallelogram $ABCD$ lowered the vertical $DR$ and $DS$ on the lines $BC$ and $AB$ . Prove that $\triangle DSR$ and $\triangle PBQ$ are similar .

Answer 1

$DS||BQ$ and $DR||BP \Rightarrow \angle PBQ= \angle RDS (1)$

$DP=BR, DQ=BS, \angle D = \angle B \Rightarrow \triangle QDP = \triangle SBR \Rightarrow PQ||SR$

$PQ||SR, BP||DR \Rightarrow \angle BPQ= \angle DRS (2) $

(1) and (2) $\Rightarrow \triangle PBQ \sim \triangle RDS$