Question

In the given figure, the point P bisects AB and DC. Prove that $\Delta$APC $\cong \Delta$ BPD.

Answer 1

$\begin{array}{c}pbisectorofABandCD\\ \therefore PA=PB\\ PD=PC\\ and\angle APC=\angle BOD\end{array}$

Hence, $\Delta APC\cong \Delta BDP$ by S.A.S criteria