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:
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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).
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