PROPOSITION 36. THEOREM.
If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square on the line which touches it.
Let D be any point without the circle ABC, and let DCA, DB be two straight lines drawn from it, of which DCA cuts the circle and DB touches it: the rectangle AD, DC shall be equal to the square on DB.
First, let DCA pass through the centre E, and join EB. Then EBD is a right angle. [III. 18.
And because the straight line AC is bisected at E, and produced to D, the rectangle AD, DC together with the square on EC is equal to the square on ED. [II. 6.
But EC is equal to EB;
therefore the rectangle AD, DC together with the square on EB is equal to the square on ED.
But the square on ED is equal to the squares on EB, BD, because EBD is a right angle. [I. 47.
Therefore the rectangle AD, DC, together with the square on EB is equal to the squares on EB, BD.
Take away the common square on EB;
then the remaining rectangle AD, DC is equal to the square on DB. [Axiom 3.
Next let DCA not pass through the centre of the circle ABC; take the centre E; [III. 1.
from E draw EF perpendicular to AC; [I. 12.
and hoin EB,EC,ED.
Then, because the straight line EF which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles, it also bisects it; [III. 3.
therefore AF is equal to FC.
