# XY is a line parallel to side BC of a triangle ABC. If BEAC and CFAB meet at E and F repectively. Show that ar(ABE) = ar(ACF)

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 )

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