Q. 84.1( 277 Votes )

# 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).

Answer :

It is given that:

In Δ ABC, XY||BC & BE||AC & CF||AB

**To prove:** Ar(Δ ABE) =Ar(Δ ACF)

**Proof:**

Let XY intersect AB & BC at points M and N respectively.

The figure of the question is:

Now, XY || BC

And, we know that the parts of the parallel lines are parallel

⇒ EN || BC .... (1)

Also, it is given that BE || AC

⇒ BE || CN .... (2)

From 1 and 2

BCNE is a parallelogram (because both the pair of opposite sides are parallel)

Now, the parallelogram BCNE & ΔABE have a common base i.e. BE & BE is parallel to AC.

⇒ ar(ΔABE) = ½ area of parallelogram BCNE ...(i)

Similarly, we BCFM is also a parallelogram.

And, the ΔACF and the parallelogram BCFM lie on the same base i.e. CF

⇒ ar(ΔACF) = ½ area of parallelogram BCFM ...(ii)

Now,

Parallelograms BCNE and BCFM are on the same base BC and between the same parallels BC and EF

hence there area will be equal.

⇒ Area (BCNE) = Area (BCFM)...(iii)

From the equations (i), (ii) and (iii), we can write that,

Ar(Δ ABE) = Ar(Δ ACF)

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