In figure (2) given below, the side AB of the parallelogram ABCD is produced to E. A straight line through A is drawn parallel to CE to meet CB produced at F and parallelogram BFGE is completed. Prove that

area of || gm BFGE = area of || gm ABCD.

Join AC and EF.

Since, ∆AFC and ∆AFE lie on same base AF and between same parallel lines AF and CE so,

area of ∆AFC = area of ∆AFE

Now, subtract area of ∆ABF on both sides,

area of (∆AFC - ∆ABF) = area of (∆AFE – ∆ABF)

area of ∆ABC = area of ∆BEF

2 area of ∆ABC = 2 area of ∆BEF …….(i)

From figure,

As || gm ABCD and ∆ABC lie on same base AB and between same parallel lines AB and DC.

Area of ∆ABC = $\dfrac{1}{2}$ Area of || gm ABCD

Area of || ABCD = 2 Area of ∆ABC ………(ii)

As || gm BFGE and ∆BEF lie on same base BE and between same parallel lines FG and BE.

Area of ∆BEF = $\dfrac{1}{2}$ Area of || gm BFGE

Area of || BFGE = 2 Area of ∆BEF ………(iii)

Substituting value from (ii) and (iii) in (i) we get,

Area of || ABCD = Area of || BFGE.

Hence, proved that area of || ABCD = area of || BFGE.