Chapter 4.2, Problem 3PSA

To determine

To Prove: $AB=CP.$

Explanation of Solution

Given information: $PC\perp AC,PB\perp AB,\angle PAC=\angle PAB$ .

Formula used: Congruency of triangle.

Proof:

  

  $\begin{array}{l}\angle ACP=\angle ABP\text{(each}\text{equal}\text{to}{\text{90}}^{\text{o}}\text{)}\\ \text{AP=AP}\text{(Reflective property)}\\ \angle PAC=\angle PAB(\text{given})\\ \therefore \Delta PAC\cong \Delta PAB(\text{AAS})\\ \therefore PB=PC\text{(CPCT)}\end{array}$