Chapter 5, Problem 9RP

To determine

To Show: $\overline{GC}\parallel \overline{ED}$ .

Explanation of Solution

Given:

  

  $\overline{AB}\cong \overline{CD},\overline{AG}\cong \overline{BE},\overline{AG}\parallel \overline{BE}$

Concept Used:

When two parallel lines are intersected by a transversal alternate interior angles are equal

Proof:

Here, we have

  

Since, when two parallel lines are intersected by a transversal alternate interior angles are equal.

  $\begin{array}{l}\overline{AB}\cong \overline{CD},\overline{AG}\cong \overline{BE},\overline{AG}\parallel \overline{BE}\\ \because \overline{AB}\cong \overline{CD}\\ \Rightarrow \overline{AB}+\overline{BC}\cong \overline{CD}+\overline{BC}\\ \Rightarrow \overline{AC}\cong \overline{BD}\\ \text{In }△GAC\text{ and }△EBD\\ \Rightarrow \overline{GA}\cong \overline{EB}\left(\because \text{ Given}\right)\\ \Rightarrow \angle EFJ=\angle HGJ\left(\because \text{ Corresponding angles}\right)\\ \Rightarrow \overline{AC}\cong \overline{BD}\\ \therefore △GAC\cong △EBD\\ \Rightarrow \angle GCA=\angle EDB\left(\because \text{ Corresponding angles}\right)\\ \Rightarrow \overline{GC}\parallel \overline{ED}\end{array}$