Theorem 9.1 - Parallelograms on same base and between same parallels

Theorem 9.1 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2
Theorem 9.1 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3

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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 : r (ABCD) = r (EFCD) Proof : Since opposite sides of parallelogram are parallel Also, AD = BC In AED and BFC DAB = CBF DEA = CFE AD = BC AED BFC AED BFC Hence, r ( AED) = r ( BFC) Now, r (ABCD) = r ( ADE) + r(EBCD) = r ( BFC) + r (EBCD) = r ( EFCD) Hence, proved

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