Definition D1
Two distinct points B, C on a circle divide it into two arcs. To each arc assign a size greater than 0° and less than 360°, with the two sizes summing to 360°. This size is called the central angle of the arc.
Proof.
(1)Write arc BC for the size of arc BC. The two arcs cut off by a diameter are called semicircles.convention
(2)For a point A on the circle not on arc BC, the angle ∠BAC is called the inscribed angle on that arc.convention
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Lemma L1
In triangle XYZ, extending side XY beyond Y to a point W, ∠ZYW = ∠YXZ + ∠XZY
Proof.
(1)Y is between X and W, so ∠XYW = 180°assumption, A4
(2)Z is not on line XY, so ray YZ is interior to ∠XYW.assumption, A6
(3)∠XYZ + ∠ZYW = ∠XYW = 180°(1), (2), A5
(4)∠YXZ + ∠XYZ + ∠XZY = 180°A3
(5)∠XYZ + ∠ZYW = ∠XYZ + (∠YXZ + ∠XZY)(3), (4), A11, E
(6)∠ZYW = ∠YXZ + ∠XZY(5), A11
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Lemma L2
If AD is a diameter and C is a point on the circle distinct from A and D, then ∠DAC = ½ arc DC (the arc DC not containing A)
Proof.
(1)OA = OC = rA1
(2)In triangle OAC, ∠OAC = ∠OCA. Call this value α.(1), A2
(3)O is between A and D, so ∠COD is an exterior angle of triangle OAC.assumption, A10
(4)∠COD = ∠OAC + ∠OCA = α + α = 2α(3), L1, (2), A11
(5)Ray AO and ray AD are the same, so ∠DAC = ∠OAC = α(3), (2)
(6)Arc DC and arc CA together form the semicircle arc DA, so arc DC + arc CA = ∠DOA = 180°(3), D1, A7, A8, A4
(7)Since arc CA > 0, arc DC < 180°(6), D1
(8)arc DC = ∠DOC(7), A8
(9)arc DC = ∠COD = 2α = 2∠DAC(8), (4), (5)
(10)∠DAC = ½ arc DC(9), A11, E
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Theorem T
For three points A, B, C on a circle, ∠BAC = ½ arc BC, where arc BC is the arc not containing A.
Proof.
(1)∠BAC is the inscribed angle on the arc BC not containing A.D1
(2)Let D be the other endpoint of the diameter through A.A10
(3)Either D equals B or C, or D lies on the arc BC not containing A, or D lies on the arc containing A. These three cases are exhaustive.(2), D1
(4)Case 1. If D = B, then AB is a diameter, so ∠BAC = ½ arc BC(3), L2
(5)The case D = C is the same as (4) after exchanging the names B and C.(4)
(6)Case 2. If D lies on the arc BC not containing A, then ∠BAD + ∠DAC = ∠BAC(3), A9
(7)AD is a diameter, so ∠BAD = ½ arc BD, ∠DAC = ½ arc DC(2), L2
(8)arc BD + arc DC = arc BC(3), A7
(9)∠BAC = ½ arc BD + ½ arc DC = ½(arc BD + arc DC) = ½ arc BC(6), (7), (8), A11
(10)Case 3. If D lies on the arc containing A and D ≠ A, then D lies on arc AB or arc AC. Suppose it lies on arc AB.(3), D1
(11)Then B lies on the arc DC not containing A, so ∠DAB + ∠BAC = ∠DAC, arc DB + arc BC = arc DC(10), A9, A7
(12)AD is a diameter, so ∠DAC = ½ arc DC, ∠DAB = ½ arc DB(2), L2
(13)∠BAC = ∠DAC − ∠DAB = ½(arc DC − arc DB) = ½ arc BC(11), (12), A11, E
(14)The case D on arc AC is the same as (10)–(13) after exchanging the names B and C.(13)
(15)In all three cases, ∠BAC = ½ arc BC(4), (5), (9), (13), (14)
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