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  • XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF)

Q: 8     XY is a line parallel to side BC of a triangle ABC. If  \small BE\parallel AC and   \small CF\parallel AB meet XY at E and F respectively, show that

              \small ar(ABE)=ar(ACF)

Answers (1)

best_answer

We have a \DeltaABC  such that BE || AC and CF || AB
Since XY || BC and BE || CY 
Therefore, BCYE is a ||gm

Now, The ||gm BCEY and \DeltaABE are on the same base BE and between the same parallels AC and BE.
\therefore ar(\DeltaAEB) = 1/2 .ar(||gm BEYC)..........(i)
Similarly, ar(\DeltaACF) = 1/2 . ar(||gm BCFX)..................(ii)

Also, ||gm BEYC and ||gmBCFX are on the same base BC and between the same parallels BC and EF.
\therefore ar (BEYC) = ar (BCFX).........(iii)

From eq (i), (ii) and (iii),  we get

ar(\Delta ABE) = ar(\DeltaACF)
Hence proved

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manish

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