/
Source Image
Loading...
English
h 15. def. 1.
k 4 def. of this.
ce the section DHE will be a circle, whose center is G, and therefore the figure ADE is a cone.
l
Coroll. 1. The straight line DE is the diameter of the circle DHE.
m 24 def. 11.
2. The cone ADE is similar to the cone ABC (because AG . GD
AF . FB.)
---
### Prop. V.
Fig. 9.
If a Scalene cone is cut by a plane (ABC) through the axis at right angles to the base (BC), and is cut by another plane (GHK) at right angles to the triangle through the axis, which cuts off from the side of the vertex (A) a triangle (AGK) similar to that (ABC) which is through the axis, but positioned subcontrarily; the section (GHK) will be a circle. And let it be called a subcontrary section.
a 11. 11.
b 38. 11.
c 3. def. 11.
d 15. 11.
e 4. of this.
f 35. 3.
g 29. 1.
h hyp.
k 15. 1.
l 4. 6.
m 16. 6.
n prior.
o conv. 35. 3.
In the section GHK let any point be taken; from which a straight line HF is drawn to the plane ABC, which will fall onto the straight line GK, and indeed perpendicularly, namely at F. Through F let DE be drawn parallel to BC; and the plane through DE and HF is parallel to the base BC; and it makes the section DHE a circle, in which FD * FE = FH sq. But because angle ADE = (angle ABC =) angle AKG, and angle KFE = GFD; the triangles KFE and DFG will be equiangular. Whence EF . FK
GF . FD. Therefore FK * GF = (EF * FD =) HF sq. Wherefore the section GHK is a circle. Q. E. D.
---
### Prop. V.
Fig. 9.
If a Scalene cone is cut by a plane (ABC) through the axis at right angles to the base (BC), and is cut by another plane (GHK) at right angles to the triangle through the axis, which cuts off from the side of the vertex (A) a triangle (AGK) similar to that (ABC) which is through the axis, but positioned subcontrarily; the section (GHK) will be a circle. And let it be called a subcontrary section.
a 11. 11.
b 38. 11.
c 3. def. 11.
d 15. 11.
e 4. of this.
f 35. 3.
g 29. 1.
h hyp.
k 15. 1.
l 4. 6.
m 16. 6.
n prior.
o conv. 35. 3.
In the section GHK let any point be taken; from which a straight line HF is drawn to the plane ABC, which will fall onto the straight line GK, and indeed perpendicularly, namely at F. Through F let DE be drawn parallel to BC; and the plane through DE and HF is parallel to the base BC; and it makes the section DHE a circle, in which FD * FE = FH sq. But because angle ADE = (angle ABC =) angle AKG, and angle KFE = GFD; the triangles KFE and DFG will be equiangular. Whence EF . FK
Prop. VI.
Fig. 10.
If a cone is cut by a plane (ABC) through the axis, and some point (D) is taken on the surface of the cone which is not on the side of the triangle through the axis; and from it a straight line (DE) is drawn equidistant to a certain straight line (MN), which is perpendicular from the circumference of the circle (BMC) to the base of the triangle (BC); it will meet the triangle through the axis, and being produced further to the other part of the surface, it will be bisected by that triangle.
a 30. 1.
b
Let the straight line AD extended meet the circumference of the base at point K; through which let KHL be drawn parallel to MN; and therefore to DE. Therefore DE produced will meet the straight line AH, namely at F. Let DF be produced to the surface at G.
It is clear that the straight line AL (since it is in the plane ADE, or AKL, and on the surface of the cone) will meet DF at G. And it will be KH . DF
(AH . AF
) HL . FG. Whence, since KH = HL, it will be DF = FG. Q. E. D.Use ← → arrow keys to navigateSwipe left/right to navigate·Produced by SourceLibrary.org in Amsterdam, 2026