# Area – State And Prove Theorems

## 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.

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

In _ ADE and _ BCF,

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

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

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

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

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

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

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

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

= ar (EFCD)

So, parallelograms ABCD and EFCD are equal in area.

## Theorem

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

## PROOF

Now, suppose ABCD is a parallelogram whose one of the diagonals is AC

Let AN⊥ DC. Note that

△ ADC ≅ △ CBA (Why?)

So, ar (ADC) = ar (CBA) (Why?)

Therefore, ar (ADC) = ½ ar (ABCD)

= ½ (DC X AN)

So, area of △ ADC = ½ × base DC × corresponding altitude AN

In other words, area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height). For having equal corresponding altitudes, the triangles must lie between the same parallels.