Question

For the following figure, can you deduce from the information $\angle ADC\cong \angle CBA$ and$\angle BAD\cong \angle DCB$ that quadrilateral ABCD is a parallelogram. If so, what theorem can you use?

Short answer

The quadrilateral$ABCD$ is a parallelogram.

Step-by-step solution

Step 1- Check the figure.

Consider the figure.

Step 2- Apply the concept of the interior angle.

If two lines are cut by a transversal line and the sum of two interior angles on the same side of the transversal is supplementary then those two lines are parallel.

Step 3- Step description.

Consider $\angle ADC\cong \angle CBA;\angle BAD\cong \angle DCB$.

Consider the angle sum property of quadrilateral as follows:

$\begin{array}{c}\angle ADC+\angle CBA+BAD+\angle DCB={360}^{\circ }\\ 2\angle ADC+2\angle DCB={360}^{\circ }\\ \angle ADC+\angle DCB={180}^{\circ }\end{array}$

Since “If two lines cut by a transversal line and sum of two interior angles on the same side of the transversal are supplementary then that two lines are parallel” thus$\overline{AD}$is parallel to $\overline{BC}$ such that $\overline{AD}\left|\right|\overline{BC}$.

Similarly,

$\begin{array}{c}\angle ADC+\angle CBA+BAD+\angle DCB={360}^{\circ }\\ 2\angle BAD+2\angle ADC={360}^{\circ }\\ \angle BAD+\angle ADC={180}^{\circ }\end{array}$

Since “If two lines cut by a transversal line and sum of two interior angles on the same side of the transversal are supplementary then that two lines are parallel” thus$\overline{AB}$is parallel to $\overline{DC}$ such that $\overline{AB}\left|\right|\overline{DC}$.

Thus, $ABCD$ is a parallelogram.