Chapter 3.5, Problem 41HP

(a)

To determine

Prove that $a\parallel c$

Explanation of Solution

Formula Used:

Converse of Corresponding Angle Theorem:

When a transversal intersects two lines , the two lines are parallel if the corresponding angles are congruent Given:

  $m\angle 1+m\angle 2=180°$

  

Calculation :

Given: $m\angle 1+m\angle 2=180°$

To Prove: $a\parallel c$

Proof:

Statement	Reason
$m\angle 1+m\angle 2=180°$	Given
$m\angle 2+m\angle 3=180°$	$\angle 2\text{ and }\angle 3$ are linear.
$m\angle 1=m\angle 3$	$\begin{array}{l}m\angle 1+m\angle 2=180°=m\angle 2+m\angle 3\\ \Rightarrow m\angle 1+m\angle 2=m\angle 2+m\angle 3\\ \Rightarrow m\angle 1=m\angle \mathrm{3.........................}[\text{subtract }m\angle 2\text{ from each side]}\end{array}$
$\angle 1\cong \angle 3$	$m\angle 1=m\angle 3$
$a\parallel c$	Converse of Corresponding Angles Theorem.

And $\angle 1\text{ and }\angle 3$ are corresponding angles which are congruent, where t is the transversal to the lines a and c .

By the Converse of Corresponding Angle Theorem which states that when a transversal intersects two lines , the two lines are parallel if the corresponding angles are congruent, the lines a and c are parallel.

So, $a\parallel c$ .

Hence proved.

(b)

To determine

Prove that $t\perp c$

Explanation of Solution

Formula Used:

Corresponding Angle Theorem:

If two lines are parallel , then corresponding angles are congruent.

Given:

  $m\angle 1+m\angle 3=180°$ and $a\parallel c$

  

Calculation :

To Prove: $t\perp c$

Proof:

Statement	Reason
$m\angle 1+m\angle 3=180°$	Given
$a\parallel c$	Given
$\angle 1\cong \angle 3$	Corresponding Angle Theorem , since $\angle 1\text{ and }\angle 3$ are corresponding angles and $a\parallel c$
$m\angle 1=m\angle 3$	$\angle 1\cong \angle 3$
$m\angle 3=90°$	$\begin{array}{l}m\angle 1+m\angle 3=180°\\ \Rightarrow m\angle 3+m\angle 3=180°\mathrm{....}[m\angle 1=m\angle 3]\\ \Rightarrow 2m\angle 3=180°\\ \Rightarrow m\angle 3=\frac{180°}{2}\mathrm{............}[divide\text{ by 2]}\\ \Rightarrow m\angle 3=90°\end{array}$
$m\angle 3+m\angle 4=180°$	They are linear.
$m\angle 4=90°$	$\begin{array}{l}m\angle 3+m\angle 4=180°\\ \Rightarrow 90°+m\angle 4=180°\mathrm{.....}[m\angle 3=90°]\\ \Rightarrow m\angle 4=180°-90°\mathrm{.....}[\text{subtract 90}°\text{]}\\ \Rightarrow m\angle 4=90°\end{array}$
$t\perp c$	$m\angle 4=90°$

Hence proved.