Chapter 4.2, Problem 1PSA

To determine

To Prove: AM is perpendicular bisector of BC.

Explanation of Solution

Given information: AM is median, AB=AC.

Formula used: Congruency of triangle.

Proof:

  

  $\begin{array}{l}AB=BC\text{(given)}\\ AM=CM(\text{reflective property})\\ BM=CM(\mathrm{AM}isMedian)\\ \therefore \Delta ABM\cong \Delta ACM(\text{SSS})\\ \text{By}\text{linear}\text{pair}\text{axiom,}\\ \angle \text{AMB+}\angle \text{AMC=}{180}^{\circ }\\ \therefore \angle \text{AMB}={90}^{\circ }\end{array}$