parallelogram are on the same base and between the same parallels are eqaul in area show that.
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Theorem: Parallelograms on the same base and between the same parallels are equal in area. ... BC = AD (opposite sides of a parallelogram are equal)
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Answer:
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 AB∥CD and ED∥FC 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
⇒△AED≅△BFC [ 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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