Chapter 4, Problem 5RP

To determine

To Prove: CD is perpendicular bisector of AB.

Explanation of Solution

Given information: $AB=BC,AD=DB,\angle MAC=\angle MAC,\angle MAD=\angle MBD.$

Formula Used: Congruency of triangle, linear pair axiom

Proof:

  

  $\begin{array}{l}AC=BC,AD=DC\text{(given)}\\ CD=CD\text{(common)}\\ \therefore \Delta ACD\cong \Delta BCD(\text{SSS})\\ \therefore \angle ACM=\angle BCM(\text{CPCT})\\ \therefore \Delta ACM\cong \Delta BCM(\text{ASA})\\ \therefore AM=BM\&\angle AMC=\angle BMC(\text{CRT})\\ \angle AMC+\angle BMC={180}^{\circ }\left(\text{Linear pair axiom}\right)\\ \therefore \angle AMC={90}^{\circ }=\angle BMC.\end{array}$