Equal corresponding angles ⇒ the two lines are parallel

Dependencies: Exterior angle > either non-adjacent interior angle (Euclid I.16, a result of neutral geometry); Protractor axiom (angle additivity).

Statement

Let line $t$ cut lines $\ell_1$, $\ell_2$ at points $P_1$, $P_2$ ($P_1\neq P_2$). The angle at $P_1$ formed by $t$ and $\ell_1$, together with the angle at $P_2$ formed by $t$ and $\ell_2$ in exactly the corresponding position (same side of $t$, same side of the respective transversal line), is called a pair of corresponding angles, denoted $\angle 1$, $\angle 2$. The theorem asserts:

By Vertical angles are equal we immediately get Equal alternate angles ⇒ parallel and Co-interior angles supp. ⇒ parallel — the three are essentially the same statement.

Proof

By contradiction. Suppose $\ell_1\not\parallel\ell_2$; then the two lines meet at some point $Q$, giving $\triangle P_1P_2Q$ ($P_1$, $P_2$ are not collinear with $Q$ outside $\ell_1\cup\ell_2$, so the three points are not collinear).

Without loss of generality, suppose $Q$ lies on one side of $t$, with $\angle 1$, $\angle 2$ being the corresponding angles on the other side of $t$. At $P_1$, $\angle 1$ and the interior angle $\angle QP_1P_2$ of the triangle form a linear pair (they piece together along $\ell_1$ into $180^\circ$), so $\angle 1$ is precisely the exterior angle of $\triangle P_1P_2Q$ at $P_1$ (on one side of $t$). At $P_2$, $\angle 2$ is the interior angle $\angle QP_2P_1$ of the triangle itself — the non-adjacent interior angle to $\angle 1$.

By the triangle exterior-angle inequality:

This contradicts the hypothesis $\angle 1 = \angle 2$. Hence $\ell_1\parallel\ell_2$. $\blacksquare$

Immediate consequences

• Alternate interior / co-interior angles: applying Vertical angles are equal once at $P_1$ flips $\angle 1$ to the other side, immediately giving "Equal alternate angles ⇒ parallel"; one more application of "linear-pair sum = $180^\circ$" gives "Co-interior angles supp. ⇒ parallel".
• Drawing a parallel through a point off the line: given a line $\ell$ and a point $P$ not on $\ell$, draw $t\perp\ell$ through $P$, then draw $\ell'\perp t$ through $P$; the corresponding angles formed by $t$ across $\ell$ and $\ell'$ are both $90^\circ$, so by this theorem $\ell'\parallel\ell$ — this gives the existence of parallel lines (without invoking the parallel postulate).
• Bridging forward and backward: this theorem is the upper limit of L2's "neutral geometry": going one step further (corresponding angles equal $\Leftarrow$ two lines parallel) requires the parallel postulate, the work of L3.

Remarks

This is "half" of Euclid I.28 — proving only the $\Rightarrow$ direction. The $\Leftarrow$ direction (Parallel ⇒ corresponding angles equal) is I.29, which must use the fifth postulate; the present section, together with Exterior angle > either non-adjacent interior angle and uniqueness of the perpendicular, all still belongs to "neutral geometry" — equally valid in hyperbolic geometry.

The pairing "$\angle 1$ is the exterior angle, $\angle 2$ is the non-adjacent interior angle" used in the proof depends on the positional definition of corresponding angles: same side of $t$, same side of the respective transversal — if $Q$ is on one side of $t$, then the corresponding angles on the other side of $t$ are exactly the "exterior at $P_1$ / interior at $P_2$" pairing. Switching to the other pair of corresponding angles (the one on $Q$'s side) swaps the roles in the proof; the conclusion is unchanged.