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)

In quadrilateral BCYE, BE ∥ CY [∵ BE ∥ AC]
BC ∥ EY [∵ BC ∥ XY]
Therefore, BCYE is a parallelogram.
Triangle ABE and Parallelogram BCYE are on the same base BE and between
same parallels, BE ∥ AC.
Hence, ar(ABE) = (1/2)ar(BCYE) …(1)
[∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.]
Similarly, triangle ACF and parallelogram BCFT are on the same base CF and Between same parallels CF ∥ AB.
Hence, ar(ACF) = (1/2)ar(BCFX) …(2)
And, ar(BCYE) = ar(BCFX) …(3)
[∵ On the same base (BC) and between same parallels (BC ∥ EF), area of parallograms are equal]
From the equation (1), (2) and (3), ar (ABE) = ar(ACF)