The Conics of Apollonius (1675) - Page 32

a that it meets the section on both sides; if not, when extended it will meet A E, let us say at E; b therefore it will meet the section sooner, let us say at G; therefore let G F be applied as an ordinate, c and let A F : A D

A D : A B; d whence F D : D B

A D : A B. Therefore, when B C is drawn parallel to G F, the square of F D : the square of D B (e this is the square of G F : the square of B C) f

the square of A D : the square of A B g

A F : A B. Therefore, since the square of G F : the square of B C

A F : A B, h point C will be on the section: wherefore C D meets the section on both sides.
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### Prop. XXVIII.
Fig. 38.

If a straight line (C D) touches one (A) of the opposite sections, and a point (E) is taken within the other section (B), and a line (E F) is drawn through it parallel to the tangent, then when extended in both directions, it will meet the section (B).

a 4. 6.
b constr.
c 14. 5.
d constr. & Sch. 48. 1.
e 7. 5.
f 21. of this.

For because C D meets the diameter, E F will meet the same, let us say at G. Let A H = B G, and through H let H K be drawn parallel to C D or E F, meeting the section at K, and let K L be applied as an ordinate, and let G M = H L be taken; finally, let M N be drawn parallel to K L. H L : L K

a G M : M N. Therefore (because H L b = G M) c it will be that L K = M N. d Also B L * A L = A M * B M. Wherefore B L * A L : the square of L K e

A M B M : the square of M N. f Whence point N will be on section B. By a similar argument, E F will meet the section on the other side.
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### Prop. XXIX.
Fig. 39.
If in opposite sections a straight line (C D) drawn through the center (C) meets one section (A), then when extended further, it will cut the other (B) as well.

a 21. of this.
b constr & sch. 48. 1.
c 14. 5.
d constr.
e 29. 1.
f 4. 1.
g Sch. 15. 1.

a Let D F be applied as an ordinate to the diameter A B, and let B G = A F; and let G E be drawn as an ordinate, and let C E be joined. B F A F : the square of D F a

T : R a :: A G * B G : the square of G E. Therefore, since b B F * A F = A G * B G, c the square of D F = the square of G E, and D F = G E. d Also C F = C G, and angle F e = angle G. f Therefore, angle F C D = G C E: g wherefore the line D C E is one straight line, meeting section B at E.

Corollary 1. D F = E G.
2. C D = C E (because the triangles C F D and C G E are similar, and the sides C F and C G are equal).