# 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:

 Statement1. ∠SPR = ∠PRQ.2. ∠SRP = ∠RPQ.3. PR = PR.4. ∆PSR ≅ ∆RQP. 5. ar(∆PSR) = ar(∆RQP). (Proved) Reason1. 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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