Every Diagonal of a Parallelogram Divides it into Two Triangles of Equal Area

Here we will prove that every diagonal of a parallelogram divides it into two triangles of equal area.

Given: PQRS is a parallelogram in which PQ ∥ SR and SP ∥ RQ. PR is a diagonal of the parallelogram.

To prove: ar(∆PSR) = ar(∆RQP).

Proof:

            Statement

1. ∠SPR = ∠PRQ.

2. ∠SRP = ∠RPQ.

3. PR = PR.

4. ∆PSR ≅ ∆RQP.

5. ar(∆PSR) = ar(∆RQP). (Proved)

            Reason

1. SP ∥ RQ and PR is a transversal.

2. PQ ∥ SR and PR is a transversal.

3. Common side.

4. By ASA axiom of congruency.

5. By area axiom for congruent figures.

Note: ar(∆PSR) = ar(∆PQR) = \(\frac{1}{2}\) × ar(parallelogram PQRS).






9th Grade Math

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