Chapter 4.3, Problem 21PPS

To determine

Find two-column proof of theorem.

Answer to Problem 21PPS

  $\angle C\cong \angle C\text{'}$

Explanation of Solution

Given information:

  $\begin{array}{l}\angle A\cong \angle A\text{'}\\ \angle B\cong \angle B\text{'}\end{array}$

Calculation:

Here, we have to find the two-column proof of theorem,

  $\angle C\cong \angle C\text{'}$

  

  

  $\begin{array}{l}Now,\\ \angle A\cong \angle A\text{'};\angle B\cong \angle B\text{'}\left(given\right)\\ m\angle A=m\angle A\text{'}andm\angle B=m\angle B\text{'}\\ Byusingtriangleanglesumtheorem:\\ m\angle A+m\angle B+m\angle C=180\\ m\angle A\text{'}+m\angle B\text{'}+m\angle C\text{'}=180\\ Bytransitiveproperty:\\ m\angle A+m\angle B+m\angle C=m\angle A\text{'}+m\angle B\text{'}+m\angle C\text{'}\\ Bysubstitution:\\ m\angle A+m\angle B+m\angle C=m\angle A+m\angle B+m\angle C\text{'}\\ m\angle C=m\angle C\text{'}\\ \angle C\cong \angle C\text{'}\end{array}$

Hence, proved.