#### theorem

**Theorem:** Parallelograms on the same base and between the same parallels are equal in area.**Proof :** Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC are given in following fig.

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

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

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

Now, ar (ABCD) = ar (ADE) + ar (EDCB)

= ar (BCF) + ar (EDCB) [From(5)]

= ar (EFCD)

So, parallelograms ABCD and EFCD are equal in area.