# Theorem 9.1

Last updated at Nov. 23, 2017 by Teachoo

Last updated at Nov. 23, 2017 by Teachoo

Transcript

Theorem 9.1 Parallelograms on the same base and between the same parallels are equal in area. Given : Two parallelograms ABCD & EFCD, that have the same base CD & lie between same parallels AF & CD. To Prove : 𝑎r (ABCD) = 𝑎r (EFCD) Proof : Since opposite sides of parallelogram are parallel Also, AD = BC In∆ AED and ∆ BFC ∠DAB = ∠CBF ∠DEA = ∠CFE AD = BC ∴ ∆AED ≅ Δ BFC ∴ ∆AED ≅ Δ BFC Hence, 𝑎r (∆AED) = 𝑎r (∆ BFC) Now, 𝑎r (ABCD) = 𝑎r (∆ADE) + 𝑎r(EBCD) = 𝑎r (∆BFC) + 𝑎r (EBCD) = 𝑎r (∆ EFCD) Hence, proved

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