Question

State true or false:

The alongside figure shows a parallelogram $ABCD$ in which $AE=EF=FC$, then
$DEBF$ is a parallelogram.

• A. True
• B. False

Answer 1

The correct option is A True

Consider $\Delta AEF$ and $\Delta EFC$

$AE=FC$

$EF=EF$

$\angle AFE=\angle FEC$

$\therefore \Delta AEF$ and $\Delta EFC$ are congruent.

Hence $AF=EC$.

We know $AB=CD\therefore AB-AF=CD-EC$ i.e. $DE=FB$

Also, $AB||CD\therefore FB||ED$

Similarly using $\Delta AFD$ and $\Delta BCE$ we can prove that $DF||EB$ and $DF=EB$

Hence, DEBF is a parallelogram.