Given ABCD is a trapezium where AB∥DC.
BD intersect CD at X and BC at Y i.e, BD∥XY.
We have to prove that ar(ADX)=ar(BDY)
For △BDX and △BDY lie on the same base BD and are between parallel lines BD and XY.
[ triangle with same base and between the same parallels are equal in area. ]
For △BDX and △ADX lie on the same base DX and are between parallel lies AB and DX.
[ As AB∥DC, parts of parallel lines are parallel ]
Triangles with same base and between the same parallel are equal in area.
Hence, the answer is proved.