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)
Since, a diagonal of a parallelogram divides it into two triangles of equal area
∴ area (ΔADC) = area (ΔABC)
⇒ area (ΔAPO) + area (11gm DLOP) + area (ΔOLC)
⇒ area ( Δ AOM) + ar (11gm 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 (11gm DLOP) = area ( 11gm BMOQ)
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- Corollary: A rectangle and a parallelogram on the same base and between the same parallels are equal in area.