Postulate 1: The angles of a straight line add up to 180 degrees.

Lemma 1: Vertical angles are congruent

Let AB and EF be two lines that intersect at E.

Then:

∠ EGB ≅ ∠ AGF

Proof:

(1) ∠ EGB + ∠ EGA = 180 degrees. [Postulate 2]

(2) ∠ EGA + ∠ AGF = 180 degrees.

(3) So ∠ EGB = 180 - ∠ EGA and ∠ AGF = 180 - ∠ EGA

(4) So, ∠ EGB ≅ ∠ AGF.

QED

Lemma 2: If any corresponding angles of an intersected line are congruent, then all corresponding angles are congruent.

Proof:

(1) We start with ∠ EGA ≅ ∠ EHC

(2) Using Postulate 1, we know that ∠ EGB ≅ ∠ EHD since:

∠ EGA + ∠ EGB = 180 so that ∠ EGB = 180 - ∠ EGA

∠ EHC + ∠ EHD = 180 so that ∠ EHD = 180 - ∠ EHC

(3) Using Lemma 1, we get:

∠ FGA ≅ ∠ FHC since:

∠ FGA ≅ ∠ EGB and ∠ FHC ≅ ∠ EHD

∠ FGB ≅ ∠ FHD since:

∠ FGB ≅ ∠ EGA and ∠ FHD ≅ ∠ EHC

QED

Since any one corresponding angle implies the rest, we can use this as a definition for parallel lines.

Definition 1: Two lines are parallel if any and only if the corresponding angles of an intersecting line are congruent.

In the case of the example above, lines AB and CD are parallel if we can prove any of the following true:

(a) ∠ EGB ≅ ∠ EHD

(b) ∠ EGA ≅ ∠ EHC

(c) ∠ FHC ≅ ∠ FGA

(d) ∠ FHD ≅ ∠ FGB

Postulate 2: From any point not on a line, it is possible to construct a line that intersects the point and is parallel to the other line.

Lemma 3: Alternate angles of parallel lines are congruent

Proof:

(1) ∠ AGF ≅ ∠ CHF [From Definition of Parallel Lines]

(2) ∠ CHF ≅ ∠ EHD [From Lemma 1: Verical Angles]

(3) ∠AGF ≅ ∠ EHD

QED

Lemma 4: The angles of a triangle add up to 180 degrees

Proof:

(1) Let EC be a line that is parallel AB [Postulate 2]

(2) ∠ ECA ≅ ∠ BAC [Lemma 3, Alternating angles of parallel lines]

(3) ∠ ECD ≅ ∠ ABC [Lemma 2, Corresponding angles of parallel lines]

(4) Now ∠ ACB + ∠ ECA + ∠ ECD = 180 degrees [Postulate 1]

(5) So ∠ ABC + ∠ BAC + ∠ ACB = 180 degrees too.

QED

References

- David Joyce, Euclid's Elements