Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.
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).
From Every Diagonal of a Parallelogram Divides it into Two Triangles of Equal Area to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.