Chapter 5.5, Problem 4PSA

To determine

To show: $\overline{DE}\cong \overline{BF}$ .

Explanation of Solution

Given information:

ABCD is a parallelogram.

  

  $\overline{AE}\cong \overline{CF}$

Calculation:

ABCD is a parallelogram.

  

  $\begin{array}{l}\text{In above figure,}\\ \text{ABCD is a parallelogram}\text{. }\\ \angle ADC\cong \angle ABC\left\{\text{Opposite angles of a parallelogram}\right\}\\ \overline{AD}=\overline{BC}\left\{\text{Opposite sides of a parallelogram}\right\}\\ \overline{DC}=\overline{AB}\left\{\text{Opposite sides of a parallelogram}\right\}\\ \Delta ABC\cong \Delta CDA\left\{\text{Two triangles with two side and one angle equal}\right\}\\ \overline{AE}\cong \overline{CF}\left\{\text{Given}\right\}\\ \text{So, }\overline{EC}\cong \overline{FA}\\ \overline{DE}\cong \overline{BF}\left\{\text{Properties of equal triangle}\right\}\end{array}$