Question

prove that " parallelograms on same base and between same parallel lines are equal in area"​

Answer 1

$\huge\red{Answer}$

$\huge\red{Explanation:}$

Example:-

$ar(ABCD)=ar(EFCD$

Hence,opposite sides of parallelogram are parallel

$AB||BCAB$

( with Transvernal)

$Angle \: DAB = Angle \: CBF$(corresponding angles)

$ED||FC$

(with Transvernal) EF

$Angle \: DEA =Angle \: CFE$

(corresponding angel )

$Here, AB=BC$

(opposite sides of parallelogram is equal)

ΔAED and ΔBEC

Angle DEA=Angel CEE

Angle DAB=Angle CBF

AB=BC

Therefore, ΔAED≈ΔBFC

Hence ΔAEO≈ΔBFC

ar(ABCD)=ar(ΔABE) +ar(EBCD)

=ar(ΔBFC) +ar(EBCD)

=ar(ΔEFCD)

proved!!!

Refer the attachment for diagram

$\huge\red{Hope-it-helps!!!}$

Answer 2

Given:- There are two parallelograms ABCD and ABEF  on the same base AB and between equal parallel lines AB and FC.

To prove:- Area of prallelogram ABCD and area of parallelogram ABEF are equal.

Proof:- We know that, opposite sides of a parallelogram are equal in lengths.

Hence,

In Parallelogram ABCD, AD = BC

And In Parallelogram ABEF, AF = BE

Again, WE know that, opposite sides of parallelogram are parallel to each other.

∴In Parallelogram ABCD , AD || BC

And, In Parallelogram ABEF, AF || BE

Then,

         ∠DAF = ∠CBE

Now,

In ΔADF and ΔBCE,

                               AD = BC

                           ∠DAF = ∠CBF

                               AF = BE

∴ From, Side-Angle-Angle  condition, ΔADF and ΔBCE are congruent

∴Area of ΔADF = Area of ΔBCE    

Now,

Area of Parallelogram ABCD = Area of ABED + Area of ΔBCE

          = Area of ABED + Area of ΔADF

          = Area of parallelogram ABEF