ABCD is a parallelogram in which E is midpoint of AD and O is a point on AC such that AO=AC/4. When EO produced meets AB at F. Prove that F is mid point of AB

construction: draw diagonal BD and mark intersection point of the diagonals as X.
RTP: F is mid point of AB
proof:
AX=XC=AC/2                            [diagonals of a triangle bisect each other]
AC=2AX
also, AO=AC/4                                  [given]
or, AO=2AX/4
or,AO=AX/2................[1]
In triangle ADX
AE=DE                                      [E is midpoint of AD]
AO=OX                                            [from 1]
therefore,EO||DX                                [By mid point theorem]
In triangle AXB
AO=OX                                             [from 1]
FO||XB                                               [since,EO||DX]
therefore,AF=FB  ….....................[2]           [by converse of mid point theorem]
Therefore, F is mid point of AB            [from 2]
(Hence proved)
Hope this helps!!!