Chapter 5.2, Problem 23WE

To determine

To proof: The DEBF is a parallelogram.

Explanation of Solution

Given:

The parallelogram is shown below:

  

DE and BF are perpendicular to AC .

Calculation:

Consider in triangles AFB and CEF.

  $\begin{array}{l}\angle BAF=\angle DCE\left(\text{Alternet}\text{interior}\text{angle}\right)\\ \angle CED=\angle AFB=90°\left(\text{Given}\right)\\ AB=CD\left(\text{Sides of parallelogrma}\right)\\ \therefore \Delta ABF\cong \Delta CDE\left(\text{AAS}\right)\\ DE=BF\left(\text{CPCT}\right)\end{array}$

Consider in triangles EBF and FDE .

  $\begin{array}{l}DE=BF\\ \angle CED=\angle AFB=90°\left(\text{Given}\right)\\ EF=EF\left(\text{Common}\right)\\ \therefore \Delta EBF\cong \Delta FDE\left(\text{SAS}\right)\\ EB=FD\left(\text{CPCT}\right)\end{array}$

Thus, The DEBF is a parallelogram as opposite sides are equal.