Chapter 4.2, Problem 14PSC

To determine

To Prove: P is midpoint of AC.

Explanation of Solution

Given information: AB=BC, PM=PN, M & N are midpoints.

Formula used: congruency of triangle.

Proof:

  

  $\begin{array}{l}\therefore \frac{1}{2}AB=\frac{1}{2}BC.\\ \therefore BM=BN.\\ PM=PN \left(given\right)\\ PB=PB \left(reflectiveproperty\right)\\ \therefore \Delta MBP\cong \Delta NBP \left(SSS\right)\\ \therefore \angle ABP=\angle CBP \left(CPCT\right)\\ BP=BP \left(reflectiveproperty\right)\\ \therefore AP=PC \left(CPCT\right)\end{array}$