**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).**

**Solution:**

It is given that ** **

** (1)**

Also, we have ** (Given)**

**(2)**

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.

** (3)**

**(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.

** (4)**

Similarly, we have and parallelogram BXFC on the same base CF and between the same parallels CF and AB.

** (5)**

From **(3)**, **(4)** and **(5)**, we can say that