Question

ABCD is a parallelogram. Through point A, a line AEF is drawn to meet BC at E. DC produced to meet at F. Show that ar (BEF)=ar (DCE).​

Answer 1

Area of ΔBEF =  Area of ΔDCE

Step-by-step explanation:

Please refer the attached picture for the diagram.

Let's join one diagonal AC.

Since ABCD is a parallelogram, we know that:

AB  =  CD  and AB  | |  CD

BC  =  DA and BC | |  DA

Area of triangle  =  1/2 * base * height.

Area of ΔABC  =  Area of ∆ ABF  (As both have same base AB and same height as they lie between the same set of parallel lines)

Area of ΔABE  +  Area of ΔAEC  =  Area of ΔABE  + Area of ΔBEF.

So Area of ΔAEC  =  Area of ΔBEF -------- (1)

Similarly Area of Δ AEC  =  Area of ΔDCE -------------(2)

(As both have same base CE and lie between the same set of parallel lines)

From equations 1 and 2 , we get:

Area of ΔBEF =  Area of ΔDCE

Hence proved.