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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

$\small ar(ABE)=ar(ACF)$

Views

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

Now, The ||gm BCEY and $\Delta$ABE are on the same base BE and between the same parallels AC and BE.
$\therefore$ ar($\Delta$AEB) = 1/2 .ar(||gm BEYC)..........(i)
Similarly, ar($\Delta$ACF) = 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($\Delta$ACF)
Hence proved

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