Question

parallelograms on the same base and between the same parallels are equal in area​

Answer 1

Step-by-step explanation:

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

hope this help u plz mark me as brainlist......

Answer 2

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge\underline\bold\orange{Question}$

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

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge\underline\bold\orange{Solution}$

$\pink{Given}$

Two ||gms ABCD and ABEF on the same base AB and between the same parallels AB and FC

$\blue{To\:prove}$

ar(||gm ABCD)= ar(||gm ABEF).

$\green{Proof}$

In ∆ADF and ∆BCE, we have:

AD = BC (opp.sides of a ||gm)

∠ADF = ∠BCE (corresponding ∠s)

DF = CE ⠀⠀∴ DC = FE (each equal to AB)

⠀⠀⠀⠀⠀⠀⠀⠀⇒DC - FC = FE - FC ⇒ DF = CE

∴ ∆ADF ≅ ∆BCE (SAS-criterion).

And so, ar(∆ADF) = ar(∆BCE)

⠀⠀⠀⠀(congruent figures have equal areas).

Now, ar(||gm ABCD) = ar(∆ADF) + ar(quad.ABCF)

⠀⠀⠀⠀=ar(∆BCD) + ar(quad.ABCF)

⠀⠀⠀⠀⠀⠀⠀=ar(||gm ABEF).

Hence, ar(||gm ABCD) = ar( ||gm ABEF).

━━━━━━━━━━━━━━━━━━━━━━━━━