Definition D1
The point P is not on the circle. A line through P meeting the circle at two points is called a secant. Suppose two distinct secants ℓ₁, ℓ₂ meet the circle at A, B and at C, D respectively.
Proof.
(1)If P is outside the circle, assign the names so that A is between P and B, and C is between P and D.A14, convention
(2)If P is inside the circle, then P is between A and B, and also between C and D.A15
(3)The only common point of the two secants is P, and P is not on the circle, so A, B, C, D are four distinct points.A5, convention
∎
Lemma L1
Relative to line BC, A lies on the same side as P. If P is outside the circle, D lies on the opposite side from P; if inside, on the same side as P.
Proof.
(1)ℓ₁ and ℓ₂ are distinct and both pass through P, so they meet only at P.D1, A5
(2)If line BC equaled ℓ₁, then C would be a common point of the two secants, so C = P; but C is on the circle, a contradiction. Hence line BC differs from both ℓ₁ and ℓ₂.(1), D1, R
(3)Two distinct lines meet in at most one point, so line BC meets ℓ₁ only at B and ℓ₂ only at C. In particular A and D are not on line BC.(2), D1, A5
(4)If P were on line BC, that line would pass through P and B and so equal ℓ₁. This contradicts (2), so P is not on line BC.(2), D1, A5, R
(5)If P is outside, A is between P and B; if inside, P is between A and B. In either case B is not on segment AP.D1
(6)Segment AP lies on ℓ₁, and line BC meets ℓ₁ only at B, so segment AP does not meet line BC. Hence A and P are on the same side.(3), (4), (5), A4
(7)If P is outside, C is between P and D, so segment PD meets line BC at C. Hence D is on the opposite side from P.(3), (4), D1, A4
(8)If P is inside, P is between C and D, so C is not on segment PD, and segment PD does not meet line BC. Hence D is on the same side as P.(3), (4), D1, A4
∎
Lemma L2
In both cases, ∠PAC = ∠PDB
Proof.
(1)A and D are on the circle and not on line BC, so they lie on the same arc if on the same side, and on different arcs if on opposite sides.L1, A3
(2)Case P inside. A and D are both on the same side as P, hence on the same arc, and the arc BC not containing A and the arc BC not containing D are the same arc.(1), L1
(3)The two inscribed angles are half of the same arc, so ∠BAC = ∠BDC(2), A1
(4)P is between A and B, so ray AP and ray AB are the same and ∠PAC = ∠BACD1, A16
(5)P is between C and D, so ray DP and ray DC are the same and ∠PDB = ∠BDCD1, A16
(6)Hence in the inside case ∠PAC = ∠PDB(3), (4), (5)
(7)Case P outside. A is on the same side as P and D on the opposite side, so the two points lie on different arcs, and the arc BC not containing A and the arc BC not containing D are two different arcs.(1), L1
(8)Those two arcs sum to 360°, so ∠BAC + ∠BDC = ½ × 360° = 180°(7), A1, A2, A12
(9)A is between P and B, so ray AP and ray AB are opposite and ∠PAC + ∠CAB = 180°D1, A7
(10)C is between P and D, so ray DP and ray DC are the same and ∠PDB = ∠BDCD1, A16
(11)Hence in the outside case ∠PAC = 180° − ∠BAC = ∠BDC = ∠PDB(8), (9), (10), E
∎
Theorem T
If two secants through P meet the circle at A, B and at C, D, then PA·PB = PC·PD
Proof.
(1)If P is inside, ray PA is opposite to ray PB and ray PC opposite to ray PD, so ∠APC and ∠DPB are vertical angles and equal.D1, A6
(2)If P is outside, ray PA equals ray PB and ray PC equals ray PD, so ∠APC and ∠DPB are the same angle.D1, A16
(3)In both cases ∠APC = ∠DPB and ∠PAC = ∠PDB, so with two pairs of equal angles △PAC and △PDB are similar under P↔P, A↔D, C↔B.(1), (2), L2, A8
(4)Corresponding sides are proportional, so PA/PD = PC/PB(3), A9, A11
(5)Lengths are greater than 0, so PA·PB = PC·PD(4), A10, A11, A12
∎
Corollary C
If P is outside the circle, T is the point of tangency of a tangent from P, and a secant through P meets the circle at A, B, then PT² = PA·PB
Proof.
(1)A is between P and B, so ray PA equals ray PB, and ray BP equals ray BA.assumption, A14, A16
(2)The angle between the tangent PT and the chord TA equals the inscribed angle on that chord, so ∠PTA = ∠ABT = ∠PBT(1), A13
(3)∠TPA and ∠BPT are the same angle.(1)
(4)With two pairs of equal angles, △PTA and △PBT are similar under P↔P, T↔B, A↔T.(2), (3), A8
(5)Corresponding sides are proportional, so PT/PB = PA/PT(4), A9, A11
(6)Lengths are greater than 0, so PT² = PA·PB(5), A10, A11, A12
∎