Definition D1
In triangle ABC, let ∠C = 90°, and set a = BC, b = CA, c = AB.
Proof.
(1)The two sides enclosing the right angle are a and b, and the side opposite the right angle is c.convention
(2)When a line t cuts two lines, the two angles lying in opposite half-planes of t are called alternate angles.convention
∎
Lemma L1
If a line t cuts two parallel lines ℓ, m, the alternate angles are equal.
Proof.
(1)Suppose t cuts ℓ at P and m at Q, and ℓ ∥ m.assumption
(2)Call the alternate angle at P α, and the one at Q β.D1
(3)There is a unique line m′ through Q making alternate angle α with t.(2), A3
(4)The alternate angles are equal, so m′ ∥ ℓ(3), A2
(5)There is only one line through Q parallel to ℓ, so m′ = m(1), (4), A1
(6)Therefore β = α(3), (5), A3
∎
Theorem T1
The three interior angles of a triangle sum to 180°.
Proof.
(1)There is a unique line ℓ through the vertex A parallel to the line BC.A1
(2)Taking two points D, E on ℓ with A between them, the straight angle at A splits as ∠DAB + ∠BAC + ∠CAE = 180°. The details of the configuration are as in Proof 49.(1), A4, A5
(3)∠DAB and ∠ABC are alternate angles that the line AB makes with the parallels ℓ and BC, so ∠DAB = ∠ABC(1), (2), D1, L1
(4)∠CAE and ∠BCA are alternate angles made by the line AC, so ∠CAE = ∠BCA(1), (2), D1, L1
(5)∠ABC + ∠BAC + ∠BCA = 180°(2), (3), (4), E
∎
Theorem T2
If two triangles agree in two angles, they agree in the third, and therefore the two triangles are similar.
Proof.
(1)Suppose ∠A = ∠A′ and ∠B = ∠B′.assumption
(2)∠A + ∠B + ∠C = 180° and ∠A′ + ∠B′ + ∠C′ = 180°T1
(3)(∠A + ∠B) + ∠C = (∠A′ + ∠B′) + ∠C′(2), A12, E
(4)∠A + ∠B = ∠A′ + ∠B′(1), E
(5)∠C = ∠C′(3), (4), A13, E
(6)The three angles agree, so the corresponding sides are proportional. What supplied the third angle was the angle sum.(1), (5), A6
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Theorem T3
In the right triangle ABC, a² + b² = c²
Proof.
(1)(∠A + ∠B) + 90° = 90° + 90°, so ∠A + ∠B = 90°T1, D1, A12, A13, E, convention
(2)∠B > 0, so ∠A < 90°; and ∠A > 0, so ∠B < 90°. Both angles are acute.(1), A14
(3)Let H be the foot of the perpendicular from C to the line AB.A7
(4)H lies between A and B, so AH + HB = AB = c. Set AH = x, HB = y.(2), (3), A8, A9, D1
(5)In triangles AHC and ACB, ∠A is common and ∠AHC = ∠ACB = 90°(3), D1
(6)They agree in two angles, so the triangles are similar and AH : AC = AC : AB, that is, x/b = b/c(5), T2, D1
(7)b > 0 and c > 0, so b² = c · x(6), A10, A12, A15
(8)In triangles BHC and BCA, ∠B is common and ∠BHC = ∠BCA = 90°, so the same argument gives a² = c · y(3), D1, T2, A10, A12, A15
(9)a² + b² = c·y + c·x = c(y + x)(7), (8), A11, E
(10)y + x = x + y = c, so a² + b² = c · c = c²(4), (9), A12, E
∎
Statement C
In a plane where the fifth postulate fails, a² + b² = c² does not hold.
(1)The equation was obtained at the end of the chain T1 → T2 → T3, and the first link of the chain is A1.T1, T2, T3
(2)That the equation fails in a plane denying A1 is not proved here and is left as a statement. The circumstances of that plane are treated in Proof 59.convention
Proof omitted.