Question

Let $ABCD$ be a parallelogram and consider its diagonal $AC$. Draw perpendiculars $BK$ and $DL$ on to $AC$. Prove that $BK=DL$.

Answer 1

Given: $ABCD$ is a parallelogram. $BK$ and $DL$ are perpendiculars drawn to diagonals $AC$.

It is given that in quadrilateral $ABCD$, $DL$ and $BK$ are perpendicular to $AC$.

$\Delta ADC$ is congruent to $\Delta ABC$ (Diagonal divides a parallelogram into two congruent triangles)

Therefore, Area $\Delta ABC=$ Area $\Delta ADC$

That is $\frac{1}{2}AC\times BK=\frac{1}{2}AC\times DL$ implies $BK=DL$.

Hence, $BK=DL$.