Question

Given: ABCD is a $▱,\overline{CD}\cong \overline{CE}$

Prove: $\angle \mathrm{A}\cong \angle \mathrm{E}$

Short answer

It is proved that $\angle \mathrm{A}\cong \angle \mathrm{E}$.

Step-by-step solution

Step 1. Apply properties of parallelogram.

Opposite angles of a parallelogram are congruent.

In $▱ABCD$, $\angle A$and $\angle BCD$are opposite angles. Therefore, $\angle A\cong \angle BCD$.

Step 2. Description of step.

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

Step 3. Isosceles triangle theorem.

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

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

Step 4. Description of step.

As $\angle A\cong \angle BCD$, $\angle BCD\cong \angle CDE$and $\angle CDE\cong \angle E$then by the transitive property of congruence $\angle A\cong \angle E$.

Hence it is proved that $\angle A\cong \angle E$.