Chapter 5.2, Problem 19PSA

To determine

To Prove: $\overline{AC}\cancel{\cong }\overline{AD}$ .

Explanation of Solution

Given:

  

  $\begin{array}{l}\angle D\cong \angle ABE\\ \overline{BE}\cancel{\parallel }\overline{CD}\end{array}$

Concept Used:

When parallel lines is intersected by a transversal alternate exterior angles are equal.

Proof:

Here, we have

  $\begin{array}{l}\angle D\cong \angle ABE\\ \overline{BE}\cancel{\parallel }\overline{CD}\end{array}$

  $\begin{array}{l}\text{In }△\text{ABE and }△A\text{DC}\\ \angle ABE\cong \angle D\left(Given\right)\\ \because \overline{BE}\cancel{\parallel }\overline{CD}\\ \Rightarrow \angle ABE\cancel{\cong }\angle C\\ \Rightarrow \angle AEB\cancel{\cong }\angle D\\ \Rightarrow \angle ACD\cancel{\cong }\angle ADC\\ \Rightarrow \overline{AC}\cancel{\cong }\overline{AD}\end{array}$