The correct option is D All of the above
In the given figure, ABCD is a parallelogram, AF = FD and BE = EC.
AF=FD=AD2
BE=EC=BC2
AD=BC
∴AF=BE and we know that AF||BE.
Similarly, FD=EC and FD||EC
Thus, AFEB and DFEC are parallelograms.
AF = FD, and the parallelograms AFEB and DFEC lie between the same parallels
Thus, ar(AFEB) = ar(DFEC)
Also, ar(ΔDEF) = 12ar(DFEC)
∴12× ar(AFEB) = ar(ΔDEF)
Also, ar(ΔABE) = 12ar(AFEB)
Thus, ar(ΔABE) = ar(ΔDEF)
ΔAED and parallelogram ABCD lie on the same base AD and between the same parallels AD and BC.
Thus, ar(ΔAED)=12× ar(ABCD)