Theorem 1: Parallelograms on the same base and between the same parallels are equal in area.

Given:

Two parallelograms ABCD and ABEF standing on the same base AB and between the same parallel lines AB and CF.

To prove:

Area of parallelogram ABCD = Area of parallelogram ABEF

Proof: (Click steps below)

S.N. Statements Reasons
1. In ΔADF and ΔBCE
i) ∠ADF = ∠BCE (A) Being opposite sides of parallelogram ABCD.
ii) ∠AFD = ∠BEC (A) Corresponding angle being AD//BC.
iii) AD = BC (S) Corresponding AF//BE
2. ΔADF ≅ ΔBCE AAS axiom.
3. Area of ΔADF = Area of ΔBCE Congruent triangles are equal in area.
4. Area of ΔADF + area of trapezium ABED = area of ΔBCE + area of trapezium ABED Adding the same trapezium on both sides.
5. Area of parallelogram ABCD = Area of parallelogram ABEF. From statement (4) (Whole part axiom)
Hence Proved.
Theorem 2: Area of a triangle is half the area of a parallelogram standing on the same base and between the same parallels.

Given:

ΔEAB and ||gm ABCD stand on the same base AB and between the same parallels AB || EC.

To prove:

Area(ΔEAB) = ½ Area(||gm ABCD)

Construction: (Click)

Draw BF || AE to meet line CD produced at F. This forms parallelogram ABFE.

Proof: (Click)

Statements Reasons
1. Area of parallelogram ABFE = area of parallelogram ABCD 1. Both being on the same base AB and between the same parallels AB and CE.
2. Area of ΔABE =
12
area of parallelogram ABFE 		2. Diagonal EB bisects the parallelogram ABFE.
3. Area of ΔABE =
12
area of parallelogram ABCD 		3. From statements (1) and (2).
Hence Proved.
Theorem 3: The area of triangles standing on the same base and lying between the same parallels are equal.

Given:

ΔABE and ΔABC are standing on the same base AB and lying between parallel lines EC and AB.

To prove:

The Area of Δ ABE = The area of Δ ABC

Construction: (Click)

Draw AD || BC. So that ABCD is a parallelogram.

Proof: (Click)

Statements Reasons
1. Area of ΔABC =
12
area of parallelogram ABCD 		Diagonal AC bisects parallelogram ABCD.
2. Area of ΔABE =
12
area of parallelogram ABCD 		ΔABE and parallelogram ABCD are standing on the same base and lying between the same parallels.
3. Area of ΔABC = area of ΔABE From statements (1) and (2) above
Hence Proved.