Question

Let $ABCD$ be a parallelogram. Let $BP$ and $DQ$ be perpendiculars respectively from $B$ and $D$ on to $AC$. Prove that $BP=DQ$.

Answer 1

It is given that in parallelogram $ABCD$, $BP$ is perpendicular to $AC$ and $DQ$ is perpendicular to $AC$.

In $\Delta ADQ$ and $\Delta CBP$,

$AD=CD$ (Opposite sides of a parallelogram)

$\angle DAQ=\angle BCP$ and $AD||BC,AC$ (Transversal alternate angles)

$\angle DQA=\angle BPC={90}^0$ (Given)

$⸫\Delta ADQ=\Delta CBP$ (SAA)

$⸫BP=DQ$ (C.P.C.T)

Hence proved.