Chapter 5.3, Problem 9PSA

To determine

To Prove: $\overline{EB}\parallel \overline{DC}$ .

Explanation of Solution

Given:

  

  $\begin{array}{l}\overline{EA}\parallel \overline{DB}\text{and }\overline{EA}\cong \overline{DB}\\ \text{B is the midpoint of }\overline{AC}.\end{array}$

Concept Used:

When parallel lines is intersected by a transversal corresponding angles are equal. Sum of linear pairs of angles is $180°$ .

Proof:

Here, we have

  $\begin{array}{l}\text{When parallel lines is intersected by a transversal corresponding angles are equal}\text{. }\\ \because \overline{EA}\parallel \overline{DB}\text{and }\overline{EA}\cong \overline{DB}\\ \text{Since, when parallel lines is intersected by a transversal corresponding angles are equal}\text{. }\\ \text{In }△EAB\text{and }△DBC\\ \overline{EA}\cong \overline{DB}\\ \angle EAB=\angle DBC\\ \overline{AB}\cong \overline{BC}\left(\because \text{B is the midpoint of }\overline{AC}\right)\\ \therefore △EAB\cong △DBC\\ \angle EBA=\angle DCB\\ \Rightarrow \overline{EB}\parallel \overline{DC}\end{array}$