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:**

1. ∠SPR = ∠PRQ. 2. ∠SRP = ∠RPQ. 3. PR = PR. 4. ∆PSR ≅ ∆RQP. 5. ar(∆PSR) = ar(∆RQP). (Proved) |
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).

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