Commentary on Proposition III.32

Back to Proposition III.32.

The converse of this theorem is true, namely that, If a straight line drawn through one extremity of a chord of a circle make with that chord angles equal respectively to the angles in the alternate segments of the circle, the straight line so drawn touches the circle. This can, as Camerer and Todhunter remark, be proved indirectly; or we may prove it, with Clavius, directly. Let BD be the given chord, and let EF be drawn through B so that it makes with BD angles equal to the angles in the alternate segments of the circle respectively.

Let BA be the diameter through B, and let C be any point on the circumference of the segment DCB which does not contain A. Join AD, DC, CB.

Then, since, by hypothesis, the angle FBD is equal to the angle BAD, let the angle ABD be added to both;

therefore the angle ABF is equal to the angles ABD, BAD.

But the angle BDA, being the angle in a semicircle, is a right angle; therefore the remaining angles ABD, BAD in the triangle ABD are equal to a right angle.

Therefore the angle ABF is right; hence, since BA is the diameter through B,

[III. 16, Por.]EF touches the circle at B.

Pappus assumes in one place (IV. p. 196) the consequence of this proposition that, If two circles touch, any straight line drawn through the point of contact and terminated by both circles cuts off segments in each which are respectively similar. Pappus also shows how to prove this (VII. p. 826) by drawing the common tangent at the point of contact and using this proposition, III. 32.