Question

ABCD and BEFG are parallelograms and CE || AG. Prove that Area (|| gm ABCD) = Area (|| gm BEFG) 

[Hint: Prove Area (ΔABC) = Area (ΔGBE)]

Answer 1

The given figure is

In the given figure ΔACG and ΔAEG are on the same base and between same parallel.

∴ Area (ΔACG) = Area (ΔAEG)

Subtracting both side area of ΔABG, we get

Area (ΔACG) – Area (ΔABG) = Area (ΔAEG) – Area (ΔABG)

⇒ Area (ΔABC) = Area (ΔGBE)

⇒ `1/2` Area (|| gm ABCD) = `1/2` Area (|| gm BEFG)  ...[∵ AC and GE are diagonals of (|| gm  ABCD) and (|| gm BEFG)]

⇒ Area (|| gm ABCD) = Area (|| gm BEFG) (Proved).