Definition D1
A line t cuts two distinct lines ℓ, m at P and Q respectively. On ℓ take two points A, A′ with P between them; on m take two points B, B′ with Q between them; on t take a point S so that P lies between S and Q; and assign the names so that A and B lie on opposite sides of line t.
Proof.
(1)t differs from both ℓ and m; if it met either at two points, there would be two lines through two points, so t meets ℓ only at P and m only at Q.assumption, A5
(2)The points A, A′, B, B′, S can be chosen as above.A6
(3)Segment AA′ crosses t at P, so A and A′ lie on opposite sides of t; for the same reason so do B and B′.(1), (2), A3
(4)∠APQ and ∠PQB are called alternate angles. They lie between the two lines and on opposite sides of t.(3), convention
(5)∠SPA and ∠PQB′ are called corresponding angles. They lie on the same side of t, one outside the two lines and one between them.(3), convention
(6)∠APQ and ∠PQB′ are called co-interior angles. They lie between the two lines and on the same side of t.(3), convention
∎
Lemma L1
The following three conditions are equivalent: the alternate angles are equal; the corresponding angles are equal; the co-interior angles sum to 180°. This equivalence is independent of whether ℓ and m are parallel.
Proof.
(1)t meets ℓ only at P, so Q is not on ℓ; and it meets m only at Q, so P is not on m.D1
(2)Hence ray PQ lies in the interior of the straight angle ∠APA′, and ray QP lies in the interior of the straight angle ∠BQB′.(1), A4
(3)∠APQ + ∠QPA′ = ∠APA′ = 180°(2), D1, A2, A1
(4)∠PQB + ∠PQB′ = ∠BQB′ = 180°(2), D1, A2, A1
(5)Ray PS is opposite to ray PQ and ray PA is opposite to ray PA′, so ∠SPA and ∠QPA′ are vertical angles. Hence ∠SPA = ∠QPA′D1, A8
(6)Suppose ∠APQ = ∠PQB.assumption
(7)Rewriting (3)(4) gives ∠QPA′ + ∠APQ = ∠PQB′ + ∠PQB; substituting ∠APQ = ∠PQB and cancelling gives ∠QPA′ = ∠PQB′(3), (4), (6), A12, E, A11
(8)Comparing with (5) gives ∠SPA = ∠PQB′. Equal alternate angles imply equal corresponding angles.(5), (7), E
(9)Conversely, if ∠SPA = ∠PQB′, then by (5) ∠QPA′ = ∠PQB′, and rewriting (3)(4) and cancelling gives ∠APQ = ∠PQB(3), (4), (5), A12, E, A11
(10)Substituting (6) into (4) gives ∠APQ + ∠PQB′ = 180°. Equal alternate angles imply co-interior angles summing to 180°.(4), (6), E
(11)Conversely, if ∠APQ + ∠PQB′ = 180°, comparing with (4) and cancelling gives ∠APQ = ∠PQB(4), E, A11
(12)The three conditions are equivalent. This block never cited A10 and never asked whether ℓ and m meet.(8), (9), (10), (11)
∎
Theorem T1
If the alternate angles are equal, the two lines are parallel.
Proof.
(1)Suppose ∠APQ = ∠PQB.assumption
(2)Suppose ℓ and m meet at a point R.assumption
(3)If R were on t, then ℓ or m would meet t at two points, giving two lines through two points. Hence R is not on t.(2), D1, A5
(4)R is not on line t = line PQ, so the three points P, Q, R form a triangle.(3), D1
(5)R lies on one side of line t.(3), A3
(6)Case: R on the same side as B. R is on m, hence on ray QB; and on ℓ, hence on ray PA′.(5), D1, A3
(7)Ray PA is opposite to ray PR, so ∠APQ is the exterior angle of triangle PQR at P, and ∠PQR = ∠PQB is a non-adjacent interior angle.(4), (6), convention
(8)∠APQ > ∠PQB(7), A9
(9)Hence ∠APQ ≠ ∠PQB, contradicting (1).(8), (1), A13
(10)Case: R on the same side as A. R is on ray PA and on ray QB′.(5), D1, A3
(11)Ray QB is opposite to ray QR, so ∠PQB is the exterior angle of triangle PQR at Q, and ∠QPR = ∠APQ is a non-adjacent interior angle.(4), (10), convention
(12)∠PQB > ∠APQ, so ∠APQ ≠ ∠PQB, again contradicting (1).(11), (1), A9, A13
(13)Both cases yield a contradiction, so ℓ and m do not meet. That is, ℓ ∥ m(5), (9), (12)
(14)This block cited only the exterior angle theorem, plane separation, and the first postulate. A10 was never used.(13), A9, A3, A5
∎
Theorem T2
If a line cuts two parallel lines, the alternate angles are equal.
Proof.
(1)Suppose ℓ ∥ m.assumption
(2)ℓ and m do not meet, and Q is on m, so Q is not on ℓ.(1), D1
(3)In the half-plane of line t on the B side there exists a unique ray QC with ∠PQC = ∠APQ.D1, A7
(4)Let m′ be the line containing ray QC. Since C and A lie on opposite sides of t, ∠APQ and ∠PQC are alternate angles of ℓ and m′.(3), D1, A6, A3
(5)The alternate angles are equal, so m′ ∥ ℓ(3), (4), T1
(6)There is exactly one line through Q parallel to ℓ, so m′ = m(1), (2), (5), A10
(7)Ray QC and ray QB both lie on line m and on the same side of t, so they are the same ray.(3), (6), D1, A3
(8)∠PQB = ∠PQC = ∠APQ(3), (7), E
∎
Corollary C1
If a line cuts two parallel lines, the corresponding angles are equal and the co-interior angles sum to 180°.
Proof.
(1)The alternate angles are equal.T2
(2)The three relations are equivalent, so ∠SPA = ∠PQB′(1), L1
(3)For the same reason, ∠APQ + ∠PQB′ = 180°(1), L1
∎
Corollary C2
If the corresponding angles are equal, or the co-interior angles sum to 180°, then the two lines are parallel.
Proof.
(1)If the corresponding angles are equal, then by the equivalence of the three conditions the alternate angles are equal.L1
(2)If the co-interior angles sum to 180°, the alternate angles are equal for the same reason.L1
(3)The alternate angles are equal, so ℓ ∥ m(1), (2), T1
(4)L1 and T1 do not cite A10, so this direction holds without the parallel postulate.(3), L1, T1
∎