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Q33

Expert-verified
Found in: Page 171

### Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279

# Find something interesting to prove. Then prove it. Answers may vary. Given: $▱\mathrm{ABCD};\angle 1\cong \angle 2$

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

See the 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 $\overline{AX}\cong \overline{CY}$.