Definition D1
In triangle ABC, set a = BC, b = CA, c = AB, and let A, B, C denote the interior angles at those vertices.
Proof.
(1)Each side is written with the lowercase letter of the opposite vertex.convention
(2)The vertex labels can be relabeled, so what is proved for A also holds for B and C.convention
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Lemma L1
Let H be the foot of the perpendicular from A to line BC. Then AH = b sin C and AH = c sin B
Proof.
(1)Such an H exists uniquely.A10
(2)AH ⊥ BC, so H = C occurs only when CA ⊥ CB, that is, when C = 90°.(1), convention
(3)If C = 90°, then AH = CA = b and sin C = 1, so AH = b × 1 = b sin C(2), A9, A21
(4)If C < 90°, then H ≠ C and H lies on ray CB, so ∠ACH = C(2), A11
(5)Triangle ACH has a right angle at H with hypotenuse CA = b, and the side opposite ∠ACH is AH, so AH = b sin C(4), A4, A2
(6)If C > 90°, then H does not lie on ray CB and ∠ACH = 180° − C(2), A11, A12
(7)AH = b sin(180° − C) = b sin C(6), A4, A2, A6
(8)In all three cases, AH = b sin C(3), (5), (7)
(9)In triangle ABH the hypotenuse is AB = c and the angle at B is B or 180° − B, so by the same argument as (2)–(8), AH = c sin B(1), D1, A2, A4, A6, A9, A11, A12, A21
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Lemma L2
Let H be the foot of the perpendicular from A to line BC. Then BH² = (a − b cos C)²
Proof.
(1)If C ≤ 90°, then H lies on ray CB.A11
(2)If C = 90°, then H = C, CH = 0, and cos C = 0, so CH = b cos C(1), A9, A22
(3)If C < 90°, triangle ACH has a right angle at H with hypotenuse b, and the side adjacent to C is CH, so CH = b cos C(1), A5, A2
(4)B and H lie on the same ray from C, so BH = |CB − CH| = |a − b cos C|(1), (2), (3), A13
(5)BH² = (a − b cos C)²(4), A15
(6)If C > 90°, then H does not lie on ray CB and ∠ACH = 180° − CA11, A12
(7)In triangle ACH, CH = b cos(180° − C) = b(−cos C) = −(b cos C)(6), A5, A2, A7, A24
(8)B and H lie on opposite rays from C, so BH = CB + CH = a + (−(b cos C)) = a − b cos C(6), (7), A14, convention
(9)BH² = (a − b cos C)²(8)
(10)In every case, BH² = (a − b cos C)²(5), (9)
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Theorem T1
In triangle ABC, a/sin A = b/sin B = c/sin C
Proof.
(1)b sin C = c sin BL1
(2)0 < B < 180°, 0 < C < 180°, so sin B > 0, sin C > 0A8
(3)Dividing both sides by (sin B)(sin C) gives (b sin C)/((sin B)(sin C)) = (c sin B)/((sin B)(sin C))(1), (2), E
(4)(b sin C)/((sin B)(sin C)) = b/sin B(3), A23
(5)(sin B)(sin C) = (sin C)(sin B), so (c sin B)/((sin B)(sin C)) = c/sin C(3), A18, A23
(6)b/sin B = c/sin C(3), (4), (5)
(7)Applying the same argument to the perpendicular from B to line CA gives c/sin C = a/sin AD1, L1, A8, A18, A23, E
(8)a/sin A = b/sin B = c/sin C(6), (7)
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Theorem T2
In triangle ABC, c² = a² + b² − 2ab cos C
Proof.
(1)Let H be the foot of the perpendicular from A to line BC.A10
(2)If H ≠ B, triangle ABH has a right angle at H with hypotenuse AB = c, so c² = AH² + BH²(1), A1
(3)If H = B, then AH = AB = c and BH = 0, so AH² + BH² = c² + 0 = c²(1), A22
(4)In either case, c² = AH² + BH²(2), (3)
(5)AH = b sin C, so AH² = (b sin C)² = b² sin²CL1, A18, A20
(6)BH² = (a − b cos C)²L2
(7)(a − b cos C)² = a² − 2a(b cos C) + (b cos C)²A16
(8)2a(b cos C) = 2ab cos C and (b cos C)² = b² cos²CA18, A20
(9)c² = b² sin²C + a² − 2ab cos C + b² cos²C(4), (5), (6), (7), (8)
(10)b² sin²C + b² cos²C = b²(sin²C + cos²C) = b² × 1 = b²A17, A3, A21
(11)c² = a² + b² − 2ab cos C(9), (10), A18, A19
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Corollary C
If C = 90°, then c² = a² + b²
Proof.
(1)cos C = cos 90° = 0A9
(2)2ab cos C = 2ab × 0 = 0(1), A22
(3)c² = a² + b² − 0 = a² + b²T2, (2), A22
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