Question

prove that parallelograms on the same base and between the same parallels are equal; in area?​

Answer 1

Hey.

Refer the below attachment!

Thanks.

Answer 2

Answer:

Step-by-step explanation:

Hey!

Proof : Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels

AF and DC are given (see Fig)

We need to prove that ar (ABCD) = ar (EFCD).

In Δ ADE and Δ BCF,

∠ DAE = ∠ CBF (Corresponding angles from AD || BC and transversal AF) (1)

∠ AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) (2)

Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)

Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)

So, Δ ADE ≅ Δ BCF [By ASA rule, using (1), (3), and (4)]

Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)

Now,adding ar (EDCB) both the sides

ar (ADE) + ar (EDCB) = ar (BCF)+ ar (EDCB)

ar (ABCD) = ar (EFCD)

So, parallelograms ABCD and EFCD are equal in area.