Kitap 3
18
But FC is equal to FB; therefore FB is also greater than FG, the less than the greater: which is impossible.
18
Therefore FG is not perpendicular to DE.
18
Similarly we can prove that neither is any other straight line except FC; therefore FC is perpendicular to DE.
18
Therefore etc. Q. E. D. the tangent, ἡ ἐφαπτομένη.
PROPOSITION 19.
19
If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the tangent, the centre of the circle will be on the straight line so drawn.
19
For let a straight line DE touch the circle ABC at the point C, and from C let CA be drawn at right angles to DE; I say that the centre of the circle is on AC.
19
For suppose it is not, but, if possible, let F be the centre, and let CF be joined.
19
Since a straight line DE touches the circle ABC, and FC has been joined from the centre to the point of contact, FC is perpendicular to DE; [III. 18] therefore the angle FCE is right.
19
But the angle ACE is also right; therefore the angle FCE is equal to the angle ACE, the less to the greater: which is impossible.
19
Therefore F is not the centre of the circle ABC.
19
Similarly we can prove that neither is any other point except a point on AC.
19
Therefore etc. Q. E. D.
PROPOSITION 20.
20
In a circle the angle at the centre is double of the angle at the circumference, when the angles have the same circumference as base.
20
Let ABC be a circle, let the angle BEC be an angle at its centre, and the angle BAC an angle at the circumference, and let them have the same circumference BC as base; I say that the angle BEC is double of the angle BAC.
20
For let AE be joined and drawn through to F.
20
Then, since EA is equal to EB, the angle EAB is also equal to the angle EBA; [I. 5] therefore the angles EAB, EBA are double of the angle EAB.
20
But the angle BEF is equal to the angles EAB, EBA; [I. 32] therefore the angle BEF is also double of the angle EAB.
20
For the same reason the angle FEC is also double of the angle EAC.
