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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 1: If there is an orthotome, ABG, and BD is parallel to its diameter or
itself a diameter, and
AG is parallel to the tangent to the orthotome at B, then AD will be equal
to DG. And if AD is equal to DG, then AG and the tangent to the orthotome
at B will be parallels.
Case where BD is parallel to the diameter. Where BD is the diameter go to Theorem 0B. This proof depends on all of Theorem 0.
(Basic Diagram for first part)
(diagram1) Let ABG be an orthotome and BD parallel to the diameter EZ of the orthotome, I say that AD = DG. (Here we give only one case, where ABG does not intersect ZE).
(diagram 3 = basic ) Let AG be extended to meet HZ at L, BD be extended to meet AZ at M, and let DN perpendicular to HZ be drawn, and let GRST be drawn parallel to AH to meet AZ at T. Since AZ, DN, BQ, and GK are all perpendicular to HZ, they are clearly parallel, and since HZ is parallel to GT, they are also perpendicular to GT.
(diagram 4) Since triangles, AZL, BQH, and GKL are similar: AZ : BQ : GK = ZL : QH : KL
(diagram 8) Thus ZL + KL = 2*NL = 2*NK + 2*KL. But ZL = ZK + LK. Hence, ZK + 2*KL = 2*NK + 2*KL. That is, ZK = 2*NK. Thus, ZN = NK.
Therefore, AD = DG.
(diagram 11) Suppose that AD = DG, I say that AG is parallel to BH.
(diagram 14) Therefore, AZ : BQ : GK = ZL : QF : KL = NK+QF : QF : KL
Taking the first and third ratios, ZE : KE = T(NK+QF) : T(KL), separando, ZK : KE = T(NK+QF) - T(KL) : T(KL)
Taking the second and third ratios, QE : KE = T(QF) : T(KL), ex aequali NK : QE = O(NK,2*OF) : T(QF) = O(NK,OF) : 1/2*T(QF) = NK : 1/2 OF.
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