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).
It is given that:
In Δ ABC, XY||BC & BE||AC & CF||AB
To prove: Ar(Δ ABE) =Ar(Δ ACF)
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)
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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