Theorem : Parallelograms on the same base and between the same parallels are equal in
Given: Two parallelograms ABCD and EFCD
To prove: ar (ABCD) = ar (EFCD).
Proof : Two parallelograms ABCD and EFCD, on the same base DC and between the
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.