Chapter 6.5, Problem 13WE

To determine

To prove: $m\angle PCA<m\angle QCB$ in two column form.

Explanation of Solution

Given information:

Given figure,

Given:

  $\overline{PA}\cong \overline{PC}\cong \overline{QC}\cong \overline{QB}$

Calculation:

Consider the following diagram:

The two columns proof is as follows:

Statements	Reasons
$\begin{array}{l}m\angle PCA+m\angle A+m\angle P=180°\\ m\angle QCB+m\angle QBC+m\angle Q=180°\end{array}$	Sum all angles in a triangle is $180°$
$m\angle P>m\angle Q$	Since $PA\cong PC\cong QC\cong QB$and $CA>CB$Applying Side-Side-Side Inequality.
$\begin{array}{l}m\angle PCA=m\angle A\\ m\angle QCB=m\angle QCB\end{array}$	By Isosceles triangle theorem
$m\angle PCA<m\angle QCB$	Substituting