Theorem 1: Diagonal of a parallelogram bisect the parallelogram. Proof | SciPi

Geometry - Area of Triangles and Quadrilaterals

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Contents:

Theorem 1

Theorem 2

Theorem 3

Theorem 4

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In this blog, we will study the theoritical proof of Theorem 1 relating to the Area of Triangles and Quadrilaterals.

Theorem 1

Diagonal of a parallelogram bisect the parallelogram.

Or,

The area of each of the triangles formed by drawing a diagonal of a parallelogram is half of the area of the parallelogram.

Given: ABCD is a parallelogram in which AC is its diagonal.

To prove: ∆ABC = ½ ABCD

Proof:

	Statements		Reasons
1. (i). (ii). (iii). (iv).	In ∆ABC and ∆ACD AB=DC (S) BC = AD (S) AC = AC (S) ∆ ABC = ∆ ACD	1. (i). (ii). (iii). (iv).	Opposite sides of a parallelogram. Same as (i). Common side S.S.S. axiom
2.	Area of ∆ABC = Area of ∆ACD	2.	Area of congruent triangles
3.	∆ ABC + ∆ ACD = ABCD	3.	Whole part axiom
4.	2 ∆ ABC = ABCD	4.	From statements (2) and (3)
5.	Area of ∆ ABC = ½ area of ABCD	5.	From statement (4)

Therefore, it is proved that the area of each of the triangles formed by drawing a diagonal of a parallelogram is half of the area of the parallelogram.

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Related questions:

Q1. Prove that the diagonals of a parallelogram bisect the parallelogram.

Q2. Prove that the area of each of the triangles formed by drawing a diagonal of a parallelogram is half of the area of the parallelogram.

Q3. In the given figure, the area of ∆ ABC is 28cm², find the area of (i) ∆ ACD and (ii) parallelogram ABCD. (Answer: (i) 28cm², (ii) 56cm²)

Fig: Q3

Q4. In the given figure, the area of parallelogram ABCD is 67cm², find the area of ∆ ACD. (Answer: 33.5 cm²)

Fig: Q4

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