ABCD is a parallelogram. X and Y are mid-points of BC and CD. Prove that `ar(triangleAXY)=3/8ar(||^[gm] ABCD)`

Explanation

Join BD, BY and AC.
Since, X is the mid-point of BC.
`:.` YX is the median for `DeltaCYB`
`:. Ar(DeltaCYX) = (1)/(2) ar(DeltaCYB)`
(`.:` median divides the triangle into two equal area)
`=(1)/(2) xx (1)/(2) ar(DeltaDBC) = (1)/(4) ar (Delta DBC)`
(`:.` median by divides the `DeltaDBC` into two equal areas
`=(1)/(4) xx (1)/(2) ar (||gm ABCD)`
(`.:` diagonal DB divides the `||gm` into two equal parts)
`= (1)/(8) ar(||gm ABCD)`....(1)
Now, `ar(DeltaABC) = (1)/(2) ar(||gm ABCD)`
(`.:` diagonal divides the `||gm` into two triangle of equal areas)
`2ar (DeltaABX) = (1)/(2) ar(||gm ABCD)`
(`.:` median AX divides the `DeltaABC` into two equal parts)
`rArr ar(DeltaABX) = (1)/(4) ar(||gm ABCD)`....(2)
Similarly, `ar(DeltaAYD) = (1)/(4) ar (||gm ABCD)`....(3)
Now, using (1), (2) and (3), we get
`ar(DeltaAXY) = ar(||gm ABCD) - [ar (DeltaABX) + ar(DeltaAYD) + ar(DeltaCYX)]`
`=ar(||gm ABCD) - ((1)/(4) + (1)/(4) + (1)/(8)) ar (||gm ABCD)`
`=ar (||gm ABCD) [2 - (5)/(8)] = (3)/(8) ar (||gm ABCD)`