Question

In the fig. $segAB\parallel segCD$ and $segAB\cong segCD$ prove that $segAC$ and $segBD$ bisect each other at point $P$.

Answer 1

Given: $segAB\parallel segCD,segAB\cong segCD$

Since, $segAB\parallel segCD$,
$\angle ABP=\angle CDP$ [alternate angles]

In $\Delta APB$ and $\Delta CPD$,
$\angle APB=\angle CPD$ [vertically-opposite angles]
$\angle ABP=\angle CDP$ [given]
$AB=DC$ [given]
Thus, $\Delta APB\cong \Delta CPD$ [By AAS congruency]

By CPCT,
$AP=CP$ and $BP=DP$

So, $P$ is the mid-point for $segAC$ and $segBD$.

Hence, $segAC$ and $segBD$ bisect each other at $P$.