Chapter 4.4, Problem 15PSB

To determine

To Verify: AC bisects BD perpendicularly.

Explanation of Solution

Given information: $\angle ADC={90}^{\circ }=\angle ABC.AB=AD.$

Formula used: linear pair axiom, congruency of triangles

Proof:

  $\begin{array}{l}\angle ADC={90}^{\circ }=\angle ABC(\text{given})\\ AB=AD(\text{given})\\ AC=AC(\text{Reflective}\text{property)}\\ \therefore \Delta ADC\cong \Delta BAE(\text{RHS})\\ \therefore \angle DAE=\angle BAE(CPCT)\\ AE=AE(\text{Reflective}\text{property)}\\ \therefore \Delta DAE\cong \Delta BAE(\text{SAS})\\ \therefore \angle DEA=\angle BEAandDE=EB(\text{CPCT})\\ \therefore \angle DEA+\angle BEA={180}^{\circ }(\text{linear pair})\\ \therefore \angle DEA={90}^{\circ }=\angle BEA.\end{array}$