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Given : A triangle ABC. XY is parallel to BC; BE is parallel to AC and CY is parallel to AB.
To Prove : ar ( ABE ) = ar ( ACF ).
Proof : IIgm EBCY and triangle ABE being on the same base EB and between the same paralllels EB and CA, we have
ar ( ABE ) = 1/2 ar ( IIgm EBCY ) ...................... ( 1 )
Again, IIgm BCFX and triangle ACF being on the same base CF and between the same parallels CF and BA, we have
ar ( ACF ) = 1/2 ar ( llgm BCFX ) ....................... ( 2 )
But, llgm EBCY and llgm BCFX being on the same base BC and between the same parallels BC and EF , we have
ar ( llgm EBCY ) = ar ( llgm BCFX ) ....................... ( 3 )
From, ( 1 ), ( 2 ), and ( 3 ) , we get
ar ( ABE ) = ar ( ACF )
Regards
Arun (askIITians forum expert)
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