Question

Find something interesting to prove. Then prove it. Answers may vary.

Given: $▱\mathrm{ABCD};\angle 1\cong \angle 2$

Short answer

It is proved that $\overline{AX}\cong \overline{CY}$.

Step-by-step solution

Step 1. Apply property of parallelogram.

The opposite sides of a parallelogram are congruent.

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

Step 2. Description of step.

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

Step 3. Description of step.

As $\angle 1\cong \angle 2$, $\overline{AD}\cong \overline{BC}$and $\angle DAC\cong \angle ACB$then by ASA postulate, $\Delta DAX\cong \Delta BCY$.

Step 4. Description of step.

As $\Delta DAX\cong \Delta BCY$, then by corresponding parts of congruent triangles, $\overline{AX}\cong \overline{CY}$.

Hence it is proved that $\overline{AX}\cong \overline{CY}$.