Chapter 5.3, Problem 1PSA

To determine

To Show: $\overline{AD}\cong \overline{BC}$ .

Explanation of Solution

Given:

  

  $\overline{AB}\cong \overline{DC},\overline{AB}\parallel \overline{DC}$

Concept Used:

When two parallel lines are intersected by a transversal:

• Alternate interior angles are equal

Proof:

Here, we have

  

  $\overline{AB}\cong \overline{DC},\overline{AB}\parallel \overline{DC}$

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

  $\begin{array}{l}\because \overline{AB}\parallel \overline{DC}\\ \Rightarrow \angle BAC=\angle DCA\left(\text{Alternate interior angles}\right)\\ \text{In }△DCA\text{and }△BAC\\ \overline{AB}\cong \overline{DC}\left(\text{Given}\right)\\ \angle DCA=\angle BAC\left(\text{Proved}\right)\\ \overline{CA}\cong \overline{AC}\left(\text{Common Side}\right)\\ \Rightarrow △DCA\cong △BAC\\ \Rightarrow \overline{DA}\cong \overline{CB}\\ \Rightarrow \overline{AD}\cong \overline{BC}\end{array}$