Question

$ABCD$ and $PQRC$ are rectangle where $Q$ is the mid-point of $AC$. Prove that $DP=PC$.

Answer 1

$ABCD$ and $PQRC$ are rectangles and $Q$ is midpoint of $AC.$

In $△ADC,$ $Q$ is midpoint of $AC$, such that $PQ\parallel AD$

$\Rightarrow$ $PQ\parallel AD$

$\therefore$ $P$ is mid-point of $DC$

$\Rightarrow$ $DP=PC$ [ By using converse of midpoint theorem ]