Kitap 3
37
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.
37
Thus the angle FED is right. [III. 18]
37
Now, since DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square on DE. [III. 36]
37
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.
37
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]
37
But the angle DEF is right; therefore the angle DBF is also right.
37
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.
37
Similarly this can be proved to be the case even if the centre be on AC.
37
Therefore etc. Q. E. D.
