Proportional segments in a circle

If chords AB and CD of a circle meet at O, then AO·OB = CO·OD. Proof: the angles BAD and BCD subtend the same arc, so they are equal; the angles AOD and COB are vertical, so triangles AOD and COB are similar by two angles, giving OD : OB = AO : CO. If from a point P outside the circle a tangent PA (A the point of tangency) and a secant meeting the circle at B and C are drawn, then PA² = PB·PC, because triangles PAB and PCA are similar. Likewise, for two secants from P, PA·PB = PC·PD. These equalities let us find an unknown segment from a product. In particular, for a chord through an inner point at distance d from the centre (apply the theorem to the diameter through that point), AP·PB = (R − d)(R + d) = R² − d², and for an outer point PA·PB = d² − R².