Here we will prove that triangles with equal areas on the same base have equal corresponding altitudes (or are between the same parallels).
Given: PQR and SQR are two triangles on the same base QR, and ar(∆PQR) = ar(∆SQC). Also, PN and SM are their corresponding altitudes.
To prove: PN = SM (or PS ∥ QR).
Construction: Join PS.
Proof:
Statement |
Reason |
1. \(\frac{1}{2}\) × QR × PN = \(\frac{1}{2}\) × QR × SM. |
1. Are of a triangle = \(\frac{1}{2}\) × base × altitude, and ar(∆PQR) = ar(∆SQR). |
2. PN = SM. |
2. Cancelling \(\frac{1}{2}\) × QR from statement 1. |
3. PN ∥ SM. |
3. PN ⊥ QR and SM ⊥ QR. |
4. PNMS is a rectangle. |
4. PMNS is a parallelogram by statements 2 and 3, and two angles are right angles. |
5. PN = SM (or PS ∥ QR). (Proved) |
5. By statement 4, PNMS is a rectangle. |
Corollary: Parallelograms with equal area on the same base have equal corresponding altitudes (or are between the same parallels).
Here, ar(parallelogram PQRS) = ar(parallelogram PQMN)
Therefore, ar(∆PRQ) = ar(∆PNQ)
Therefore, RN ∥ PQ. But RS ∥ PQ, NM ∥ PQ.
Therefore, RN ∥ RS and RN ∥ NM
Having common point (R or N), all the lines are coincident.
Therefore, the parallelogram have equal altitudes.
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