The correct option is
A True
Here, two parallelograms
ABCD and
EFCD, that have the same base
CD and lie between same parallels
AF and
CD.
Since opposite sides of parallelogram are parallel
AD∥BC with transversal AB
⇒ ∠DAB=∠CBF [ Corresponding angles ] ----- ( 1 )
ED∥FC with transversal EF
⇒ ∠DEA=∠CFE [ Corresponding angles ] ----- ( 2 )
In △AED and △BFC
⇒ ∠DEA=∠CFE [ From ( 2 ) ]
⇒ ∠DAB=∠CBF [ From ( 1 ) ]
⇒ AD=BC [ Opposite sides of parallelogram are equal ]
∴ △AED≅△BFC [ AAS congruence property ]
⇒ ar(△AED)=ar(△BFC) [ Area of congruent figures are equal ]
Adding ar(EBCD) on both sides,
⇒ ar(△AED)+ar(EBCD)=ar(△BFC)+ar(EBCD)
⇒ ar(ABCD)=ar(EFCD)
∴ The parallelogram based upon the same base and between the same parallel lines are equalinarea