# Prove that parallelograms on the same base and between same parallels lines are equal in area

$\\ \text{Let ABCD and EFCD be two parallelograms lying on the same}\\ \text{base CD and between the same parallels AF and CD.}\\ \text{To prove:- area of the parallelograms ABCD and EFCD are equal.}\\ \ln\ \triangle \mathrm{ADE}\ and\ \Delta \mathrm{BCF}\\ \angle \mathrm{DAE}=\angle \mathrm{CBF}\ \text{(Corresponding angles)}\\ \angle \mathrm{AED}=\angle \mathrm{BFC}\ \text{(Corresponding angles)}\\ \mathrm{AD}=\mathrm{BC}\ (Opposite sides of the parallelogram ABCD)$

$\\ \therefore \triangle \mathrm{ADE} \cong \triangle \mathrm{BCF} (By ASA congruence rule)\\ We know that congruent figures have the same area. \\ \therefore area (\Delta \mathrm{ADE})=\operatorname{area}(\Delta \mathrm{BCF}) \ldots(1) \\ Now, we have: \\ area (\mathrm{ABCD})=\operatorname{area}(\Delta \mathrm{ADE})+ area (\mathrm{EDCB}) \\ Or, area (\mathrm{ABCD})= area (\Delta \mathrm{BCF})+ area (\mathrm{EDCB}) = area (EDCF) \\ Thus, the areas of the two parallelograms \mathrm{ABCD} and \mathrm{EFCD} are the same.$

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