Chapter 4, Problem 6RP

To determine

To Prove: $BD\perp AC.$

Explanation of Solution

Given information: $AB=BC,\angle 1=\angle 2.$

Formula Used: Congruency of triangle, properties of triangle.

Proof:

  $\begin{array}{l}AB=BC,\angle 1=\angle 2(\text{given})\\ AD=DC \left(\text{sides opposite to equal angles are equal}\right)\\ BD=DF\text{(common)}\\ \therefore \Delta ABD\cong \Delta CBD\text{(SSS)}\\ \therefore \angle ADE=\angle CDE\text{(CPCT)}\\ \therefore \Delta ADE\cong \Delta CDE\text{(ASA)}\\ \therefore \angle AED=\angle CED(\text{CPCT})\\ \text{By linear pair axiom}\text{.}\\ \angle AED+\angle CED={180}^{\circ }\\ \therefore \angle AED=\angle CED={90}^{\circ }\end{array}$