Chapter 5.1, Problem 33WE

To determine

To proof: Angle 1 is equal to angle 2.

Explanation of Solution

Given:

The parallelogram is shown below:

  

ABCD is a parallelogram.

Calculation:

Line segment AD is parallel to line BC as ABCD is parallelogram.

  $\angle 3=\angle 4\left(\text{Alternate interior angle}\right)\mathrm{......}\left(1\right)$

In triangle CDE , side AC is common.

  $AC=AC$

Subtract XY from above equation as follows:

  $\begin{array}{l}AC-XY=AC-XY\\ AX=CY\end{array}$

Now, in triangle AXD and CYB ,

  $\begin{array}{l}AX=CY\\ \angle 3=\angle 4\\ AD=BC\left(\text{parallelogram}\right)\\ \therefore \Delta AXD\cong \Delta CYB\left(\text{SAS}\right)\\ \angle ADX=\angle CBY\\ or,\\ \angle 1=\angle 2\end{array}$

Hence, this proves that the angle 1 is equal to angle 2.