Question

ABCD is a parallelogram and E ks the mid point of side BC. if DE and AB when produced meet at F prove that AF =2AB

Answer 1

from triangle BEF and CED it is clear that angle BEF and angle CED is equal
also CD is parallel to BF so angle CDE is equal to angle BFE
and E is the mid point of BC
so CE is equal to EB


so CD is equal to BF
as ABCD is a parallelogram then AB is equal to CD

so BF is equal to AB
hence AF is equal to 2 times of AB

Answer 2

IN ∆S DEC AND BEF
,angleCDE=EBF
angle BEF=CED
BE=EC
SO, ∆DEC~~∆BEF
SO, BF= CD
AB+BF=AB+AB
OR,AF=2AB