Question

E is mid-point of AB, L is passing through E and L is parallel to BC . And CM is parallel to BM.....
Prove that AEF is congruent to CDF......

Answer 1

E is the mid point of AB so, AE=BE

and F is also mid point so AF=FC

Now, In ∆AEF and ∆CDF

AE=DC

EF=FD

angle AFE=angle CFD[VOA]

By SAS rule ∆AEF congruent to ∆CDF

HOPE IT WILL HELP YOU OUT.