Chapter 5.3, Problem 28PSC

To determine

To Prove: $\overline{AB}\cancel{\cong }\overline{CD}$ .

Explanation of Solution

Given:

  

  $\overline{BC}\simeq \overline{AD}$

Concept Used:

When two parallel lines are intersected by a transversal:

• Alternate interior angles are equal

Proof:

Here, we have

  

  $\overline{BC}\simeq \overline{AD}$

Join AC, we get

  

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

  $\begin{array}{l}\because \overline{CD}\parallel \overline{AB}\\ \Rightarrow \angle DCA=\angle BAC\left(\text{Alternate interior angles}\right)\left(1\right)\\ \text{In }△DCA\text{and }△BAC\\ \angle DCA=\angle BAC\left(\text{Proved}\right)\\ \overline{BC}\simeq \overline{AD}\\ \Rightarrow △DCA\simeq △BAC\\ \text{Since, similarity of triangles not assured for congruence}\text{.}\\ \Rightarrow \overline{DC}\cancel{\cong }\overline{BA}\\ \Rightarrow \overline{CD}\cancel{\cong }\overline{AB}\end{array}$