Question

prove that parralelogram on the same base and between the parallels are equal​

Answer 1

Answer:

Theorem: Parallelograms on the same base and between the same parallels are equal in area. ... BC = AD (opposite sides of a parallelogram are equal)

Step-by-step explanation:

parallelogram on the same base and the between the same parallels are equal in area

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Answer 2

Answer:

$\huge\tt\orange{ANSWER}$

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Step-by-step explanation:

Given

Two Parallelogram ABCD and EFCD lies on the same base CD and Lie between the same parallel AF and CD

We need to prove that

ar(ABCD) = ar(EFCD)

Since opposite sides of the Parallelogram is equal therefore AB || CD and ED || FC with transversal AB

∠DAB = ∠CBF [ corresponding angle ]

with transversal EF

∠DEA = ∠CEF [ corresponding angle]

AD = BC [ opposite side of parallelogram]

in ∆AED £ ∆BFC

∠DAB = ∠CBF

∠DEA = ∠CFE

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Therefore AD = BC

∆AED = ∆ BFC { by AAS}

hence, ar ( AED ) = ar ( BFC )

area of congurant fig is equal

ar( ABCD ) = ar( ∆AED ) + ar( EBCD )

= ar(∆BFC) + ar(EBCD)

= ar(EBCD)

Hence Proved

ar ( ABCD ) = ar ( EFCD )

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