Chapter 4.1, Problem 9PSB

To determine

To Prove: $\Delta ABC$ is isosceles.

Explanation of Solution

Given information:

  $AB=AC,AD\perp BC.$

Formula used: congruency of triangle.

Calculation:

  $\begin{array}{l}AB=AC,AD\perp BC(given)\\ \therefore \angle ABC=\angle ACB,\angle ADB=\angle ADC={90}^{\circ }\\ \Delta ABD\cong \Delta ACD(AAS)\\ \therefore BD=DC(CPCT).\\ DE=DE(reflectiveproperty)\\ \Delta BED\cong \Delta CAD(SAS).\\ \therefore BE=EC(CPCT).\end{array}$