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r this is t 9. 5:
v hyp.
x 3. ax. 1.
wherefore X A = Q B. Likewise C A v = C B. xTherefore C X = C Q; rthis is H G = H V. Which was the second part.
Corollary.
Thus D E is the other diameter, conjugate to the prior one, A B.
Scholium.
Fig. 22.
We shall establish the calculation more briefly:
{ B C, or C A = d.
Let { A N = r.
{ C X, or H G = a.
{ B X = d + a.
It is therefore { A X = d - a.
Moreover, 2 d : r
d - a : r/2 - ra/2d; wherefore (d - a) (A X) * r/2 - ra/2d; this is dr/2 - raa/2d = G X sq. or H C sq. Item: because 2 d : r (B A : A N)
d (B C) : r/2, it will be d (B C) * r/2, i.e., dr/2 = D C sq. Therefore D C sq. - H C sq. (i.e., H E * H D) = raa/2d. Furthermore, because D C sq. : B C sq. (this is dr/2 : dd) D E sq. : A B sq.
D E : D F (because D E, A B, D F are in continued proportion) H E : H L
H E * H D : (raa/2d) * H L * H D dr/2 : dd
r/2 : d raa/2d : aa. It will be H L H D = aa = H G sq.
By a similar line of reasoning, it will be H L H D = H V sq. whence H G = H V.
### Prop. XVI.
Fig. 23.
If through a point (C), which divides the transverse side (A B) of opposite sections into two equal parts, a certain straight line (C D) is applied as an ordinate, it will be a diameter of the sections, conjugate to the prior diameter (A B).
a 13. of this.
b 34. 1.
c 1. 6.
d 4. 6.
e 14. of this. and
f before.
g 14. 5.
h 14. 6.
Let any straight line G H parallel to A B meet the sections at points G, H, from which GK, H L are applied as ordinates; let A E, B F be the straight sides of the sections; and let the joined lines A F, B E be produced; and let K M, L N be drawn parallel to A E, B F. And A K K Mc = (G K sq.b = H L sq.a =) B L L N. Item: A K K M : A K K Bd
(K M : K Bd By a similar line of reasoning, it will be H L H D = H V sq. whence H G = H V.
### Prop. XVI.
Fig. 23.
If through a point (C), which divides the transverse side (A B) of opposite sections into two equal parts, a certain straight line (C D) is applied as an ordinate, it will be a diameter of the sections, conjugate to the prior diameter (A B).
a 13. of this.
b 34. 1.
c 1. 6.
d 4. 6.
e 14. of this. and
f before.
g 14. 5.
h 14. 6.
Let any straight line G H parallel to A B meet the sections at points G, H, from which GK, H L are applied as ordinates; let A E, B F be the straight sides of the sections; and let the joined lines A F, B E be produced; and let K M, L N be drawn parallel to A E, B F. And A K K Mc = (G K sq.b = H L sq.a =) B L L N. Item: A K K M : A K K Bd
A E : A Be
B F : B Ad L N : L Ac
) B L * L N : B L * L A. Therefore (because A K * K Mf = B L * L N) it will be that A K * K B = B L * L A.h Wherefore K B : B L :: L A : A K. And by compounding, K LUse ← → arrow keys to navigateSwipe left/right to navigate·Produced by SourceLibrary.org in Amsterdam, 2026