As Archimedes merely states the theorem, the proof does not pretend to be a reconstruction..

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We use the notation of Dijksterhuis for the square on a straight-line and rectangle formed from two straight lines. T(AB) is the square on straight-line AB, while O(AB,CD) is the rectangle formed from straight-lines AB and CD (or lines equal to AB and CD).

Theorem 3: If there is an orthotome ABG, and BD is parallel to the diameter or is itself a diameter, and certain lines parallel to the tangent to the orthotome at B are led out, then as BD is to BZ so is AD in power to EZ.

Case where BD is parallel to the diameter. Where BD is the diameter go to Theorem 0C. This proof depends on all of Theorem 0.

Lemma: Let there be two ordinates, AS, BH, to the principal diameter SQ of an orthotome, ABQ, and the tangent to one of them, BK, with SQ extended to K, and the parallel, AM, to the tangent of BK as intersecting the extension of the main diameter, SQM. Then T(AS-BM) : 2*O(AS,BM) = MK : SM.

Let there be an orthotome ABQ with diameter QS. We first prove a lemma about any two ordinates, AS, BH, and the tangent BK of one of them BH, where BQ is the tangent and with the diameter extended to K, so that BQ = QH. Let ordinate AS > BH. Join KM.

We use these facts from Theorem 0 C and Theorem 0 D.
T(AS) : T(BH) = SQ : QK,
QH = QK,
Also, triangle ASM is similar to BHK, since BK is parallel to AM and A to BH.

T(AS) : T(BH) = SQ : QK
Componendo, T(AS) + T(BH) : T(BH) = SQ + QK : QK

Also, AS : BH = SM : HK.
Hence, O(AS, BH) : T(BH) = SM : 2*QK
so that 2*O(AS, BH) : T(BH) = SM : QK

Hence, ex aequali, T(AS) + T(BH) : 2*O(AS, BH) = SQ + QK : SM = SK : SM
Separando, T(AS) + T(BH) - 2*O(AS, BH) : 2*O(AS, BH) = SK - SM : SM
Hence, T(AS - BH) : 2*O(AS, BH) = MK : SM


Theorem

Parabola basic symptomLet there be an orthotome ABG with vertex B and BD parallel to diameter, and let AD, NZ be ordinates of the orthotome. I say that T(AD) : T(NZ) = DB : BZ.

Let QS be the (principal) diameter of orthotome ABGQ, with Q the vertex, And let there be drawn AS, NX, BH ordinates to SQ. Extend the tangent and the ordinates to the extension of SQ, namely BK, NZEL, ADGM.

(Diagram 1) By the lemma,
(1) T(AS - BH) : 2*O(AS, BH) = MK : SM
and
(2) T(NX - BH) : 2*O(NX, BH) = LK : XL

Also, AS : NX = SM : XL
Hence,
(3) 2*O(AS, BH) : 2*O(NX, BH) = SM : XL

Ex aequali (from 1 and 3),
(4) T(AS - BH) : 2*O(NX, BH) = MK : XL.
Ex aequali (from 2 and 4),
(5) T(AS - BH) : T(NX - BH) = MK : LK.

(Diagram 2) First MK = BD and LK = BZ, since BK is also parallel to DM. forming two parallelograms, BKMD and BKLZ.

(Diagram 1) Secondly,
T(AS - BH) : T(NX - BH) = T(AD) : T(NZ).
For, since BH = PS, AS - BH = AP, and, since RX = BH, NX - BH = NR. Hence,
AS - BH : NX - BH = AP : NR.

But triangle APD is similar to NRZ, since AD is parallel to NZ and NX to AS. Thus,
AS - BH : NX - BH = AP : NR = AD : NZ = DG : ZE (since AD = DG and NZ = ZE).

Hence, T(AS-BH) : T(NX - BH) = T(AD) : T(NZ) = T(DG) : T(ZE).

Hence, T(AD) : T(NZ) = T(DG) : T(ZE) = BD : ZB.

Q.E.D.

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