A point `O` inside a rectangle `A B C D` is joined to the vertices. Prove that the sum of the areas of a pair of opposite triangles so formed is equal to the sum of the areas of other pair of triangles.

Explanation

Through O, draw `MN bot PQ`
`:.` area of `Delta OPQ + " area of " Delta ORS`
`= (1)/(2) xx PQ xx OM + (1)/(2) xx RS xx ON`
`= (1)/(2) xx PQ xx OM + (1)/(2) xx PQ xx ON`
(`:. RS = PQ`, opposite sides of a rectangle)
`= (1)/(2) xx PQ xx (OM + ON) = (1)/(2) xx PQ xx MN`
`= (1)/(2) xx PQ xx QR = (1)/(2) xx` area of rectangle PQRS ....(1)
Again
area of `Delta OPS + " area of " Delta OQR = " area of " square PQRS - ( " area of " Delta OPQ + " area of " Delta ORS)`
= area of `square PQRS - (1)/(2) xx` area of `square PQRS` [from (1)]
`= (1)/(2) xx " area of " square PQRS` ...(2)
From (1) and (2), we get.
area of `Delta OPQ + " area of " Delta ORS = " area of " Delta OPS + " area of " Delta OQR`