Chapter 6.3, Problem 32PPS

To determine

Prove that ABCD is a parallelogram.

Explanation of Solution

Given :

$\overline{AC}\text{ and }\overline{BD}$ bisect each other so that $\overline{BE}\cong \overline{ED}$ and $\overline{AE}\cong \overline{EC}$

Calculation:

Given :

$\overline{BE}\cong \overline{ED}$ and $\overline{AE}\cong \overline{EC}$

$\angle BEA\cong \angle DEC$ and $\angle BEC\cong \angle DEA$ , since vertically opposite angles are congruent.

So, $\Delta BEC\cong \Delta DEA$ and $\Delta BEA\cong \Delta DEC$ , by SAS Postulate.

Hence, $\angle BCA\cong CAD$ and $\angle BAC\cong DCA$ .

They are alternate interior angles , considering lines BC , AD and AC as transversal ; and considering lines AB ,DC and AC as transversal.

So, by Converse of Alternate Interior Angle Theorem , $\overline{BC}\parallel \overline{AD}$ and $\overline{AB}\parallel \overline{DC}$ .

So, ABCD is parallelogram.

Hence proved.