Question

pls solve question no 26

Answer 1

It is given that:
In Δ ABC, XY||BC &  BE||BC & 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 
Also, it is given that BE || AC
⇒  BE || CN
Therefore, 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
⇒ Area (BCNE) = Area (BCFM)...(iii)
From the equations (i), (ii) and (iii), we can write that, 
Ar(Δ ABE) =Ar(Δ ACF)