prove that diagonal of a parallelogram divide into two Triangles of equal area
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Given: A parallelogram ABCD in which BD is one of its diagonals, side AB ∥ CD and BC ∥ AD.
To Prove: Area of ∆ABD = area of ∆BCD
Proof: In ∆ABD and ∆BCD,
AD = BC [opp. Sides of a ∥ gm]
AD = CB [opp. Sides of a ∥ gm]
BD = BD [common side]
∴ ∆ABD ∆CDB [By SSS congruent Rule]
∴ Area of ∆ ABD = area of ∆BCD [congruent area Axiom]
To Prove: Area of ∆ABD = area of ∆BCD
Proof: In ∆ABD and ∆BCD,
AD = BC [opp. Sides of a ∥ gm]
AD = CB [opp. Sides of a ∥ gm]
BD = BD [common side]
∴ ∆ABD ∆CDB [By SSS congruent Rule]
∴ Area of ∆ ABD = area of ∆BCD [congruent area Axiom]
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