Q: 8 XY is a line parallel to side BC of a triangle ABC. If and meet XY at E and F respectively, show that
We have a ABC such that BE || AC and CF || AB
Since XY || BC and BE || CY
Therefore, BCYE is a ||gm
Now, The ||gm BCEY and ABE are on the same base BE and between the same parallels AC and BE.
ar(AEB) = 1/2 .ar(||gm BEYC)..........(i)
Similarly, ar(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.
ar (BEYC) = ar (BCFX).........(iii)
From eq (i), (ii) and (iii), we get
ar( ABE) = ar(ACF)