In figure, `CD || AE` and `CY || BA`. Prove that `ar (DeltaCBX) = ar (DeltaAXY)`.

Explanation

Given A` triangle ABC,` PQ||BC, BX||CA and CY||BA.
TO PROVE `ar(triangle ABX)=ar(triangleACY)`
PROOF ||gm XBCQ and `triangle`ABX being on the same bas XB and between the same parallels XB and CA, we have
`ar(triangleABX)=(1)/(2)ar("||gmXBCQ")" "...(i)`
Again, ||gm BCYP and `triangle`ACY being on the same base CY and between the same parallels CY and BA, we have
`ar(triangleACY)=(1)/(2)ar("||gmBCYP"). " "...(ii)`
But, ||gm XBCQ and ||gm BCYP being on the same base BC and between the same parallels BC and XY, we have
`ar("||gm XBCQ)=ar("||gm BCYP")." "...(iii)`
`therefore` from (i), (ii) and (iii), we get
`ar(triangleABX)=ar(triangleACY)`.