Question

GEHF is a parallelogram.

Answer 1

Given: parallelogram $ABCD$, E and F are mid points of $AB$ and $CD$ respectively. DE and EC meet AF and BF respectively at G and H
Now, In $△HEB$ and $△HFC$
$\angle EHB=\angle FHC$ (vertically opposite angles)
$EB=FC$ (Half lengths of equal sides)
$\angle HBE=\angle HFC$ (Alternate angles for parallel sides AB and CD)
Thus, $△HEB\cong △HCF$ (AAS rule)
Thus, $HE=HC$ and $HF=BH$ (By CPCT)
or H is mid point of EC and BF
Now, In $△AFB$
E is mid point of AB and H is mid point of BF.
Thus, $EH\parallel AF$ or $EH\parallel GF$ (1)
Also, F is mid point of DC and H is mid point of BF
Thus, $FH\parallel GE$ (2)
Thus, from (1) and (2)
$GEHF$ is a parallelogram.