therefore the angle ABE is equal to the angle FDE.
But the angle FDE is a right angle; [Construction.
therefore the angle ABE is a right angle. [Axiom 1.
And EB is drawn from the centre; but the straight line drawn at right angles to a diameter of a circle, from the extremity of it, touches the circle; [III. 16, Corollary.
therefore AB touches the circle.
And AB is drawn from the given point A. q.e.f.
But if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF at right angles to DE; then DF touches the circle. [III. 16, Cor.
PROPOSITION 18. THEOREM.
If a straight line touch a circle the straight line drawn from the centre to the point of contact shall be perpendicular to the line touching the circle.
Let the straight line DE touch the circle ABC at the point C; take F, the centre of the circle ABC, and draw the straight line FC: FC shall be perpendicular to DE.
For if not, let FG be drawn from the point F perpendicular to DE, meeting the circumference at B.
Then, because FGC is a right angle, [Hypothesis.
FCG is an acute angle; [I. 17.
and the greater angle of every triangle is subtended by the greater side; [I. 19.
therefore FC is greater than FG.
But FC is equal to FB; [I. Definition 15.
therefore FB is greater than FG, the less than the greater;
which is impossible.
Therefore FG is not perpendicular to DE.
In the same manner it may be shewn that no other straight line from F is perpendicular to DE, but FC;
therefore FC is perpendicular to DE.
Wherefore, if a straight line &c. q.e.d.
