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)

XY || BC = EY || BC ...(Given)
Also, BE || AC = BE || CY
\(\therefore\) we can say that, EBYC is a parallelogram.
Similarly,
XY || BC = XF || BC
Also, FC || AB = FC || XB
\(\therefore \) BCFX is a parallelogram.


Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC and EF.

\(\therefore \) Area (EBCY) = (\(\frac{1}{2} \) ) Area (BCFX) ...(i)

Now, considering parallelogram EBYC and \(\triangle{AEB}\),

We can say that, these lie on the same base BE and are between the same parallels BE and AC.

\(\therefore \) Area (\(\triangle{AEB}\)) = (\(\frac{1}{2} \) ) Area (EBYC) ...(ii)

Also, parallelogram BCFX and \(\triangle{ACF}\) are on the same base CF and between the same parallels CF and AB.

Thus, Area (\(\triangle{ACF}\)) = (\(\frac{1}{2} \) ) Area (BCFX) ...(iii)

So, from eq.(i), (ii) and (iii), we get,

Area (\(\triangle{AEB}\)) = (\(\triangle{ACF}\))
Hence, proved.