Chapter 5, Problem 12CT

To determine

To prove: that $ABCD$ is a parallelogram.

Explanation of Solution

Given information:

Given figure,

  

Given:

  $\angle ADC\cong \angle CBA$ and $\angle BAD\cong \angle DCB$

Calculation:

Consider the angle sum property of quadrilateral as,

  $\angle ADC+\angle DCB+\angle CBA+\angle BAD=360°$

Consider the following values,

  $\begin{array}{c}\angle ADC+\angle DCB+\angle CBA+\angle BAD=360°\\ 2\angle ADC+2\angle DCB=360°\\ \angle ADC+\angle DCB=180°\\ \overline{AD}\parallel \overline{BC}\end{array}$

Use converse of interior angle as below,

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

  $\begin{array}{c}\angle BAD+\angle ADC=180°\\ \overline{AB}\parallel \overline{DC}\end{array}$

Therefore, it is proved that $ABCD$ is a parallelogram.