Chapter 4.2, Problem 8PSB

To determine

To Prove: $\Delta ABC$ is isosceles

Explanation of Solution

Given information: AM is median and $AM\perp BC.$

Formula used: Congruency of triangles.

Calculation:

  

  $\begin{array}{l}AM\perp BC\text{And}\text{Am}\text{is}\text{median}\text{(Given)}\\ \therefore \angle AMB=\angle AMC(\text{Each}\text{equal}\text{to}{90}^{\circ })\\ BM=MC\\ AM=AM(\text{Reflective}\text{property})\\ \therefore \Delta AMB\cong \Delta AMC(\text{SAS})\\ \therefore \Delta ABM\cong \Delta ACM(\text{CPCT})\\ \text{Which}\text{proves}\text{that}\text{ΔABC}\text{is}\text{isosceles}\text{.}\end{array}$