Chapter 2.6, Problem 15WE

To determine

To Proof: The proof for the given table.

Explanation of Solution

Given Data:

The given statement is that $\angle 1$ and $\angle 5$ are supplementary.

The angle $\angle 3$ and $\angle 5$ are supplementary.

The given table is shown in Table 1

Table 1

$\angle 1$ and $\angle 5$ are supplementary $\angle 3$ and	?
$\begin{array}{l}m\angle 1+m\angle 5=180\\ m\angle 3+m\angle 5=180\end{array}$	?
$m\angle 1+m\angle 5=m\angle 3+m\angle 5$	?
$m\angle 5=\_\_\_\_m\angle 5$	Reflexive Prop
$m\angle 1=m\angle 3$ or   $\angle 1\cong \angle 3$	?

Calculation:

Since, $\angle 1$ and $\angle 5$ are supplementary and the angle

  $\angle 3$ and $\angle 5$ are supplementary. Then,

  $\begin{array}{l}\angle 1+\angle 5=\angle 3+\angle 5\\ \angle 1\cong \angle 3\end{array}$

Hence Proved.

The complete table is shown in Table 2

Table 2

$\angle 1$ and $\angle 5$ are supplementary$\angle 3$ and	$\angle 3$ and $\angle 5$ are supplementary.
$\begin{array}{l}m\angle 1+m\angle 5=180\\ m\angle 3+m\angle 5=180\end{array}$	Since, these angle form the supplementary pairs.
$m\angle 1+m\angle 5=m\angle 3+m\angle 5$	Since, LHS and RHS both are supplementary angle pair
$m\angle 5=m\angle 5$	Reflexive Prop
$m\angle 1=m\angle 3$ or $\angle 1\cong \angle 3$	Subtract $m\angle 5$ from both the sides in $m\angle 1+m\angle 5=m\angle 3+m\angle 5$