Prove that parllelogram on the same base and between the same parallel are equal in area
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given: two parallelograms ABCD and bcfe are on the same base BC and and lie between the same parallel ae and bc
to prove:ar (parallelogram ABCD)=ar (parallelogram bcef)
proof: in ∆abf and∆dce,
angle baf =angle cde (corresponding angles ad ab || dc and ae is thetransversal)
angle afb =angle dec -(2) (corresponding angle as fb||ec and ae is transversal)
fb=ec-(3) ( opposite sides of parallelogram)
therefore from (1) ,(2) and(3) we get
∆afb=~∆dec (by AAS congruence)
*ar ∆afb=ar ∆dec
adding ar.fbcd on both sides ,we get
ar ∆afb+ ar.fbcd =ar.∆dec +ar.fbcd
= ar ||gm abcd=ar ||gm bcef
Hence ,proved.
hope this will help you
and if you have any confusion than you can ask in comment box
given: two parallelograms ABCD and bcfe are on the same base BC and and lie between the same parallel ae and bc
to prove:ar (parallelogram ABCD)=ar (parallelogram bcef)
proof: in ∆abf and∆dce,
angle baf =angle cde (corresponding angles ad ab || dc and ae is thetransversal)
angle afb =angle dec -(2) (corresponding angle as fb||ec and ae is transversal)
fb=ec-(3) ( opposite sides of parallelogram)
therefore from (1) ,(2) and(3) we get
∆afb=~∆dec (by AAS congruence)
*ar ∆afb=ar ∆dec
adding ar.fbcd on both sides ,we get
ar ∆afb+ ar.fbcd =ar.∆dec +ar.fbcd
= ar ||gm abcd=ar ||gm bcef
Hence ,proved.
hope this will help you
and if you have any confusion than you can ask in comment box
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