Question

In the below figure, ABCD is parallelogram. O is any point on AC. PQ || AB and LM ||
AD. Prove that ar (||gm DLOP) = ar (||^gm BMOQ)

Answer 1

Since, a diagonal of a parallelogram divides it into two triangles of equal area
∴ area (ΔADC) = area (ΔABC)

⇒ area (ΔAPO) + area (11^gm DLOP) + area (ΔOLC)

⇒ area ( Δ AOM) + ar (11^gm DLOP) + area (ΔOQC) ..... (1)

Since, AO and OC are diagonals of parallelograms AMOP and OQCL respectively.
∴ area   (ΔAPO) = area (ΔAMO)       ........ (2)

And, area  (ΔOLC)  = Area  (ΔOQC)        ......(3)
Subtracting (2) and (3) from (1), we get

Area  (11^gm DLOP)  = area ( 11^gm BMOQ)