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Prop. 5



Statement (explanatory diagram) or (single general diagram) or (general diagrams with cases separated)

Let there be a segment ABG enclosed by a straight line and a section of a right-angled cone, and from the middle of AG let there be drawn BD 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.

Theorem 4 Theorem 4

Proof (explanatory diagram) or (single general diagram) or (general diagrams with cases separated)

(diagram 1) For let KH be drawn through H parallel to KH.  (diagram 2) Therefore, as BD s in length to BK, so is DG in power to KH.  For this was proved (Prop. 3 and Prop. 1, if needed).  (diagram 3) Therefore, as BG is in length to BI so will BG be in power to BQ.  For DZ and KH are equal.(*1*) (diagram 4) Therefore lines BG, BQ, BI are proportional.(*2*) (diagram 5) Thus, BG has the same ratio to BQ which GQ has to QI.(*3*)  (diagram 6) Therefore, as GD is to DZ, so is QZ to QH.(*4*)  (diagram 7) But DA is equal to DG (Prop. 1).  And so it is clear that DA has the same ratio to DZ which ZQ has to QH.

Theorem 4 Theorem 4

 

(*1*) 1. BD : BK = T(DG) : T(KH)
2. BD : BK = BG : BI (triangle BKI is similar to BDG)
3. DG : KH = DG : ZD (since KH = DZ)
4. DG : ZD = BG : BQ (since QZ is parallel to BD, and line parallel to the base of a triangle, BD, cut the sides of triangles proportionally, cf . Euclid, Elements VI 2)
5. Hence, DG : KH = BG : BQ (from 3 and 4)
6. Hence, T(DG) : T(KH) = T(BQ) : T(BI) (from 5)
7. Hence, BG : BI = T(BG) : T(BQ) (from 2, 6)

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Theorem 4Theorem 4

(*2*) T(a) : T(b) = a : c iff a : b = b : c. Here is a quick proof: T(a) : T(b) = O(a,b) : O(c,b), so that, alternando, T(a) : O(a,b) = T(b) : O(c,b). Simplying each ratio, a : b = b : c.
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(*3*) BG : BQ = BQ : BI
For the case where Z is on AD, use Euclid, Elements V 12, BG : BQ = BG + BI : BQ + BI = QG : QI.
For the case where Z is on DG, given that a : b = b : c and a > b, it follows that a : b = a - b : b -c
proof:
1. a : b = b : c
2. a : a - b = b: b - d (separando)
3. a : b = a - b: b - d (alternando)
Hence,
BG : BQ = BG - BQ : BQ - BI = QG : QI.
Of course one may prove trivially a more general principle: a : b = c : d and a > c => a : c = a - c : c - d
proof:
1. a : b = c : d
2. a : c = b = d (alternando)
3. a : a - c = b : b - d (separando)
4. a : b = a - c: b - d (alternando)
It is impossible to tell whether Archimedes has in his tool box some principle such as one of these or intends the diligent and astute reader to figure all this out. He does, after all, have a macho style.

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Theorem 4Theorem 4

(*4*) BG : BQ = GD : DZ (cf. note 1, step 4 or Euclid, Elements VI 2)
QG : QI = QZ : QH (because triangle IQH and QZG are similar).

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Theorem 4Theorem 4

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