Equal alternate interior angles ⇒ the two lines are parallel

Dependencies: Corresponding/alternate angles ⇔ lines parallel, Vertical angles are equal.

Statement

Let line $t$ cut lines $\ell_1$, $\ell_2$ at distinct points $P_1$, $P_2$. At $P_1$, $t$ forms two angles with $\ell_1$; the angle at $P_2$ that lies on the opposite side of $t$ and on the opposite side of the transversal forms with the angle at $P_1$ a pair of alternate interior angles — denote them $\angle 1$ (at $P_1$) and $\angle 2$ (at $P_2$). The theorem asserts:

Proof

Step 1: At $P_1$, flip $\angle 1$ into its vertical angle $\angle 3$.

By Vertical angles are equal (Vertical angles are equal), $\angle 1$ equals the angle $\angle 3$ at $P_1$ obtained by extending both $t$ and $\ell_1$ to the opposite side:

Step 2: $\angle 3$ and $\angle 2$ are now corresponding angles; invoke the corresponding-angles criterion for parallel lines.

Note that now $\angle 3$ (at $P_1$, on the "upper side of $\ell_1$ + right side of $t$") and $\angle 2$ (at $P_2$, on the "upper side of $\ell_2$ + right side of $t$") sit in identical positions relative to the transversal $t$ and their respective lines — they are precisely a pair of corresponding angles. Combining the hypothesis $\angle 1 = \angle 2$ with Step 1:

That is, the corresponding angles $\angle 3$ and $\angle 2$ are equal; by Corresponding/alternate angles ⇔ lines parallel (Corresponding/alternate angles ⇔ lines parallel) we conclude $\ell_1 \parallel \ell_2$. $\qquad\blacksquare$

Immediate consequences

• Co-interior angles supp. ⇒ parallel: if the interior angles on the same side at $P_1$ and $P_2$ sum to $180^\circ$, then by "linear-pair sum = $180^\circ$" we may replace the interior angle at $P_1$ with its linear-pair partner — the latter forms with the interior angle at $P_2$ a pair of equal alternate interior angles, and the present theorem yields parallelism.
• "Z-shape" parallel test: drawing a Z (or reverse Z), the two horizontal strokes are crossed by the diagonal, with the angles in the "hooks" at top and bottom forming a pair of alternate interior angles; equality of these two angles locks the horizontal strokes parallel.
• Bridging forward and backward: together with Corresponding/alternate angles ⇔ lines parallel, the three (corresponding / alternate / co-interior) say the same thing — they convert into one another via a single Vertical angles are equal or Linear pair sums to 180°.

Remarks

The proof uses Vertical angles are equal only once: turn $\angle 1$ into $\angle 3$, which is corresponding to $\angle 2$, then invoke the previous theorem. The whole step is "position transform + known criterion" — no new geometric object is introduced.

This conclusion, along with corresponding angles ⇒ parallel and Co-interior angles supp. ⇒ parallel, still belongs to "neutral geometry" (no parallel postulate required). Its converse ($\ell_1\parallel\ell_2 \Rightarrow$ alternate interior angles equal) does require the fifth postulate, and is left to L3.