Chapter 5.2, Problem 25WE

To determine

To Prove: that $AECF$ is a parallelogram.

Explanation of Solution

Given

  $ABCD\text{and }BEDF$ are parallelogram.

And given figure as:

  

Proof:

In triangles $\Delta DAE\&\Delta BCF$ .

  $\overrightarrow{AD}\cong \overrightarrow{BC}\therefore$ ( $ABCD$ is a rhombus)

  $\angle ADC\cong \angle ABC\therefore$ ( $ABCD$ is a rhombus)

  $\angle ADE\cong \angle CBF\therefore$ ( $\angle EDC\&\angle ABF$ are equal angle of parallelogram made of parallel sides two parallelogram)

ABCD is a parallelogram.

  $\angle ADC\cong \angle ABC$

  $\angle EDC\&\angle ABF$ are equal angles od parallelogram made of parallel sides.

  $\angle ADE\cong \angle CBF$

  $\overrightarrow{DE}\cong \overrightarrow{BF}\therefore \left(BEDF\text{is a}\text{parallelogram}.\right)$

Therefore, by SAS postulate $\Delta DAE\cong \Delta BCF$ .

By CPCT, $\overrightarrow{AE}\cong \overrightarrow{FC}$ .

In triangles $\Delta AFB\&\Delta CDE$ .

  $\overrightarrow{DE}\cong \overrightarrow{BF}\therefore \left(BEDF\text{is a parallelogram}\right)$

Since, $\angle EDC\&\angle ABF$ are equal angles of parallelogram made of parallel sides.

  $\angle ADE\cong \angle CBF$

  $\overrightarrow{DC}\cong \overrightarrow{BA}\therefore \left(BEDF\text{is a parallleogram}\right)$

Therefore, by SAS postulate $\Delta AFB\cong \Delta CED$ .

By CPCT, $\overrightarrow{AF}\cong \overrightarrow{CE}$ .

Since, two pairs of opposite sides $\overrightarrow{AE}\cong \overrightarrow{FC}$ and $\overrightarrow{AF}\cong \overrightarrow{CE}$ in $AECF$ are parallel therefore $AECF\text{ is a parallleogram}$ .