For let DE be drawn touching ABC;
let the centre of the circle ABC be taken, and let it be F; let FE, FB, FD be joined.
Thus the angle FED is right. [III. 18]
Now, since DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square on DE. [III. 36]
But the rectangle AD, DC was also equal to the square on DB;
therefore the square on DE is equal to the square on DB;
therefore DE is equal to DB.
And FE is equal to FB;
therefore the two sides DE, EF are equal to the two sides DB, BF;
and FD is the common base of the triangles;
therefore the angle DEF is equal to the angle DBF. [I. 8]
But the angle DEF is right;
therefore the angle DBF is also right.
And FB produced is a diameter;
and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16, Por.]
therefore DB touches the circle.
Similarly this can be proved to be the case even if the centre be on AC.