Given: A parallelogram ABCD whose one of the diagonals is BD.
To prove: ar (∆ABD) = ar (∆CDB).
Proof: In ∆ABD and ∆CDB.
AB = DC [Opp. sides of a gm ]
AD = BC [Opp. sides of a gm ]
BD = BD [Common side]
∴ ∆ABD ≅ ∆CDB [By SSS]
∴ ar (∆ABD) = ar(∆CDB) [Congruent area axiom]
