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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)

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.