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Question

Prove the theorem : Parallelogram on the same base and between the same parallels are equal in area.

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Solution

Given two parallelogram ABCD and EFCD that have the same base CD and lie between same parallel AF and CD.
We have prove that ar(ABCD)=ar(EFCD)
Since opposite sides of gm are parallel ABCD and EDFC with transversal AB
DAB=CBF [ Corresponding angles ]
with transversal EF
DEA=CFE [ Corresponding angles ]
AD=BC [ Opposite sides of parallelogram are equal ]
In AED ξ BFC
DEA=CFE
DAB=CBF
AD=BC
AEDBFC [ AAS congruency ]
Hence, ar(AED)=ar(BFC)
( Areas of congruent figures are equal )
ar(ABCD)=ar(ADE)+ar(EBCD)
=ar(BFC)+ar(EBCD)
=ar(EBCD)
ar(ABCD)=ar(EBCD)
Hence, the answer is proved.

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