Triangles with Equal Areas on the Same Base have Equal Corresponding Altitudes
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:
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Statement |
Reason |
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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). |
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2. PN = SM. |
2. Cancelling \(\frac{1}{2}\) × QR from statement 1. |
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3. PN ∥ SM. |
3. PN ⊥ QR and SM ⊥ QR. |
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4. PNMS is a rectangle. |
4. PMNS is a parallelogram by statements 2 and 3, and two angles are right angles. |
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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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