Question

Que: 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(ΔABE) = ar(ΔAC)

Answer 1

Answer:

Solution:

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) = ½ ar(BCYE) … (1)

Now,

CF || AB and XY || BC

⇒ CF || AB and XF || BC

⇒ BCFX is a parallelogram

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

∴, ar (ΔACF)= ½ ar (BCFX) … (2)

But,

||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…..(4)

Also,

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

⇒ ar(△AEB) = ½ ar(BEYC) ……(5)

Similarly,

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

⇒ ar(△ ACF) = ½ ar(BXFC) ……..(6)

From (4), (5) and (6),

ar(△AEB) = ar(△ACF)

Hope it will be helpful :)☺️

Answer 2

HEY MATE YOUR ANSWER IS IN THE GIVEN ATTACHMENT

HOPE IT IS HELPFUL

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