Given: A ∆ABC in which XY || BC, BE || CA and CF || BA.
=> BE || CY and CF || BX
To prove: ar (∆ABE) = ar (∆ACF)
Proof: ||gm EBCY and ∆ABE are on the same base BE and between the same parallels BE & CA.
ar (∆ABE) = (1/2) ar (||gm EBCY) …(1)
Again ||gm BCFX and ∆ACF are on the same base CF and between the same parallels CF and BA.
=> ar (∆ACF) = (1/2) ar (||gm BCFX) …(2)
But ||gm EBCY and ||gm BCFX are on the same base BC and between the same parallels BC and EF.
=> ar (||gm EBCY) = ar (||gm BCFX) …(3)
From (1), (2) and (3)
ar(∆ABE) = ar(∆ACF)
