Given

XY || BC, BE || AC and CF || AB

To show

ar(ΔABE) = ar(ΔAC)

Proof:

BCYE is a || gm as ΔABE and ||gm BCYE are on the same base BE and between the same parallel lines BE and AC.

∴ ar(ABE) =  \(\frac{1}{2}\) ar(BCYE) … (1)

CF || AB and XY || BC

⇒ CF || AB and XF || BC

⇒ BCFX is a || gm

As ΔACF and || gm BCFX are on the same base CF and in-between the same parallel AB and FC .

∴,ar (ΔACF)= \(\frac{1}{2}\)  ar (BCFX) … (2)

||gm BCFX and || gm BCYE are on the same base BC and between the same parallels BC and EF.

∴,ar (BCFX) = ar(BCYE) … (3)

From (1), (2) and (3), we get

ar (ΔABE) = ar(ΔACF)

⇒ ar(BEYC) = ar(BXFC)

As the parallelograms are on the same base BC and in-between the same parallels EF and BC–(iii)

△AEB and ||gm BEYC are on the same base BE and in-between the same parallels BE and AC.

⇒ ar(△AEB) = \(\frac{1}{2}\) ar(BEYC) — (iv)

Similarly,

△ACF and || gm BXFC on the same base CF and between the same parallels CF and AB.

⇒ ar(△ ACF) = \(\frac{1}{2}\) ar(BXFC) — (v)

From (iii), (iv) and (v)

ar(△ABE) = ar(△ACF)