Question

Given: ABCDis a $▱,\angle \mathrm{A}\cong \angle \mathrm{E}$

Prove: localid="1637661691185" $\overline{AB}\cong \overline{CE}$

Short answer

It is proved that $\overline{AB}\cong \overline{CE}$.

Step-by-step solution

Step 1. Apply property of parallelogram.

The opposite sides of a parallelogram are congruent.

In $▱ABCD$, $\overline{AB}$and $\overline{CD}$are opposite sides. Therefore, $\overline{AB}\cong \overline{CD}$.

Step 2. Description of step.

From the given figure, it can be observed that $\overline{BC}\parallel \overline{AD}$and $\overline{AD}$is a transversal then $\angle A$and $\angle CDE$are corresponding angles such that, $\angle A\cong \angle CDE$.

Step 3. Description of step.

It is given that, $\angle A\cong \angle E$and from step 2, $\angle A\cong \angle CDE$then by the transitive property of congruence, $\angle CDE\cong \angle E$.

Step 4. Converse of isosceles triangle theorem.

If two angles of a triangle are congruent then the sides opposite to those angles are congruent.

Consider $\Delta CDE$, in which $\angle CDE\cong \angle E$then by the converse of isosceles triangle theorem,$\overline{CD}\cong \overline{CE}$.

Step 5. Description of step.

As $\overline{AB}\cong \overline{CD}$and $\overline{CD}\cong \overline{CE}$then by the transitive property of congruence, $\overline{AB}\cong \overline{CE}$.

Hence it is proved that $\overline{AB}\cong \overline{CE}$.