Question

Ques 15 plz be quick....​

Answer 1

we know that ABCD is a parallelogram and E and F are points on the sides DC and BC .

we have to prove , ar (AED) = ar (AFD) = 1/2 ar(ABCD)

we observe that triangle AFD  and parallelogram ABCD lie on the same base AD and these are between the same parallels AD and BC.

So , ar(AFD) = 1/2 ar (ABCD)...........(1)

similarly,in triangle AEB and parallelogram ABCD lie on the same base AB and between the same parallels AB and DC.

So ,ar (AEB) = 1/2 ar(ABCD)..............(2)

From equation (1) and (2) we have

ar (AFD) = ar (AEB) =1/2 ar (ABCD)

Hence proved

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