Chapter 4.4, Problem 6PSA

To determine

To Prove: $BD=CE$

Explanation of Solution

Given information: AG is perpendicular bisector of BC and DE.

Formula used: congruency of triangles.

Proof: AG is perpendicular bisector of BC & DE (given)

  $\begin{array}{l}\therefore BF=CF,DG=GE,\angle ABF=\angle ACF=\angle ADG=\angle AEG={90}^{\circ }\\ AF=AF(\text{Reflection}\text{property)}\\ \therefore \Delta ABF\cong \Delta ACF(\text{SAS})\\ \therefore \angle BAF=\angle CAF(\text{CPCT})\\ AB=AC(\text{CPCT})\\ \text{Similarly}AD=AE.\\ \therefore AD-AB=AE-AC(\text{Subtraction property})\\ \therefore BD=CE.\end{array}$