XY is a line parallel to side BC of a triangle ABC. If and meet XY at E and F respectively, show that ar(ABE)=ar(ACF).
It is given that
Also, we have (Given)
From (1) and (2), we can say that BEYC is a parallelogram.
(A quadrilateral is a parallelogram if both the pairs of opposite sides are parallel.)
Similarly, we can show that BXFC is a parallelogram.
Therefore, we have BEYC and BXFC as two parallelograms on the same base BC and between the same parallels EF and BC.
(Parallelograms on the same base and between the same parallels are equal in area.)
We can also see that and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.
Similarly, we have and parallelogram BXFC on the same base CF and between the same parallels CF and AB.
From (3), (4) and (5), we can say that