Orthotome

Statement
Let there be a segment ABG enclosed by a straight line and an orthotome, and from the middle of AG let BD be drawn parallel to the diameter or itself a diameter, and let straight line BG be joined and extended. If in fact some other line ZQ parallel to BD intersects the straight line through points B, G, then ZQ will have the same ratio to QH which DA has to DZ.
Proof
For let KH be drawn through H parallel to KH.  Therefore, as BD is in length to BK, so is DG in power to KH.  For this was proved.  Therefore, as BG is in length to BI so will BG be in power to BQ.  For DZ and KH are equal.  Therefore lines BG, BQ and, BI are proportional.  Thus BG has the same ratio to BQ which GQ has to QI.  Therefore, as GD is to DZ, so is QZ to QH.  But DA is equal to DG.  And so it is clear that DA has the same ratio to DZ which ZQ has to QH.
Lemma 1:  BD : BK = BG : BI
(by similar triangles BDG and BKI)
Lemma 2:  BG : BQ = DG : KH

= DG : DZ
(by similar triangles BDG and QYB with YB = KH = DZ)
 --Archimedes mentions DZ = KH, but what is needed is the parallel to KH a the intersection of the HZ and BG
Lemma 3:  BQ : BI = GQ : QI (for BQ : BI = BG : BQ = BQ+-QG : BI+-IQ = QG : QI, or a : b = c : d <=> a : b = a+-c : b+-d)
Lemma 4:  QZ : QH  = GQ : QI
(by similar triangles QZG and QHI)
Lemma 5:  GQ : QI  =QZ : QH
(by similar triangles QZG and QHI)

1.    BD : BK = DG2 : KH2 (by theorem 3)
2.    Since DZ = KH,
3.    BG : BI = BG2 : BQ2 (BG : BI = BD : BK, by Lemma 1; and DG : KH = BG : BQ, by Lemma 2, and a : b = c : d <=> a2 : b2 = c2 : d2) <=> a)
Hence,
4.    BG : BQ = BQ : BI (a : b = a2 : c2 <=> a : c = c : b, since a : b = a2 : c2 <=> b : a = c2 : a2 <=> ba :  a2 = c2 : a2 <=> ba = c2 <=> a : c = c : b)

5.    BG : BQ = GQ : QI  (since BQ : BI = GQ : QI, by Lemma 3)

6.    GD : DZ = QZ : QH (since BG : BQ = GD : DZ by Lemma 2; and QZ : QH  = GQ : QI, by lemma 4)
7.    Since DA = DG,
8.    DA : DZ = ZQ : QH