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 2: If there is an orthotome ABG, and BD is parallel to the diameter or is itself a diameter, and ADG is parallel to the tangent of the orthotome at B, and EG is tangent to the orthotome at G, then BD and BE will be equal.
Case where BD is parallel to the diameter. Where BD is the diameter go to Theorem 0D. This proof depends on all of Theorem 0.

(Basic Diagram = diagram 2)

(diagram 1) Let there be the segment of an orthotome ABG with base AG and vertex B and let LQZ be tangent to the orthotome at G. Also, let BD be parallel to diameter EQ, Extend DB to meet tangent ZG at Z. I say that ZB = ZD.

(diagram 2) Let tangent BX at B meet EQ extended at X and ZG to meet QE at L. Let ordinates BM, and GK be drawn perpendicular to EQ. Clearly, LE = EK, and XE = EM, by Theorem 0D. Also draw ZN perpendicular to XQ. Note: the proof strategy is to show that since BD = XP (trivial), and MN = BZ (trivial) and ME = EX (trivial) and EN = EP (most of the work), BD = XP = EX - EP = ME - NE = MN = BZ.

(diagram 3) Clearly, as BDPX and BZNM are parallelograms, XP = BD and ZB = NM and BM = ZN.

(diagram 4) By similar triangles, ZN : GK = NL : KL, but ZN = BM, so that BM : GK = NL : KL. (diagram 5) So too by similar triangles, BM : GK = MX : KP (note that BX is parallel to DP),
and ZN : GK = NL : KL.
Thus, BM : GK = MX : KP = NL : KL.
(diagram 6) By Theorem 0C and D, T(BM) : T(GK) = ME : KE = 2*ME : 2*KE = MX : KL
Hence, MX : KL = T(BM) : T(GK) = T(MX) : T(KP), so that MX : KP = KP : KL.
(diagram 7) Similarly, MX : KL = T(MX) : T(KP) = T(NL) : T(KL) , so that MX : NL = NL : KL.
(diagram 8) Therefore, NL = KP (as means between the same magnitudes), so that PL = NK (remove KL from each). And since ME = EX and EK = EL, therefore ZB = NM = ME - (KN+EK) = EX - (EL + PL) = XP = XD.


(diagram 9) Suppose DB = BZ. I say that LZ is tangent to the segment at G. Suppose not. Then draw a tangent through G (by Theorem 0D) so as to meet DZ, extended if need be, at T. Then, RT is tangent to the segment at G and BT = BD and is unequal to BZ, although BZ = DB, which is impossible.