Point $P$ in parallelogram $ABCD$ is such that $\angle APB+\angle CPD=180^\circ$. Prove $AP\cdot CP+BP\cdot DP=AB\cdot BC$

Question

$P$ is any point inside a parallelogram $ABCD$ such that $\angle APB +\angle CPD=180^\circ$. Prove that $$AP\cdot CP + BP\cdot DP = AB\cdot BC$$

Sum of areas are equal. Also tried to apply similarity but it didn't worked.

Answer 1

Let $APQD$ be a parallelogram.

Thus, $PBCQ$ is a parallelogram, $\Delta ABP\cong\Delta DCQ$ and $\measuredangle DQC+\measuredangle DPC=180^{\circ},$

which says that $PCQD$ is cyclic.

Id est, by Ptolemy $$DQ\cdot CP+CQ\cdot DP=CD\cdot PQ$$ or $$AP\cdot CP + BP\cdot DP = AB\cdot BC.$$