Co-interior angles supplementary ⇒ parallel

Dependencies: Corresponding/alternate angles ⇔ lines parallel, Linear pair sums to 180°.

Statement

Let line $t$ cut lines $\ell_1$, $\ell_2$ at points $P_1$, $P_2$ ($P_1\neq P_2$). On a single side of $t$, the angle $\angle 1$ formed at $P_1$ by $t$ and $\ell_1$, together with the angle $\angle 2$ formed at $P_2$ by $t$ and $\ell_2$, is called a pair of co-interior angles — they lie between the two transversal lines, on the same side of $t$. The theorem asserts:

Proof

We only need Linear pair sums to 180° and Corresponding/alternate angles ⇔ lines parallel. Flip $\angle 1$ along $\ell_1$ to the other side of $P_1$, and it coincides with $\angle 2$ in the corresponding-angle position.

Step 1: at $P_1$, flip $\angle 1$ along $\ell_1$ into the linear-pair angle $\angle 3$.

At $P_1$, let $\angle 3$ denote the angle that pairs with $\angle 1$ along $\ell_1$ to form a linear pair (it lies on the opposite side of $t$). By Linear pair sums to 180°:

Step 2: substitute the hypothesis to merge $\angle 3$ and $\angle 2$ into a corresponding-angles equality.

Substituting the hypothesis $\angle 1 + \angle 2 = 180^\circ$ (i.e. $\angle 2 = 180^\circ - \angle 1$):

By the positional definition of "corresponding angles": $\angle 3$ lies on the side of $t$ opposite to $\angle 1$ and on the same side of $\ell_1$ as before; together with $\angle 2$ at $P_2$ (on the same side of $t$ and on the same side of its respective transversal line), they form precisely a pair of corresponding angles. Equal corresponding angles, by Corresponding/alternate angles ⇔ lines parallel, give $\ell_1 \parallel \ell_2$. $\blacksquare$

Immediate consequences

• The three criteria are essentially identical: Corresponding/alternate angles ⇔ lines parallel, Equal alternate angles ⇒ parallel, and co-interior angles supplementary ⇒ parallel all derive from one another by an algebraic shuffle of a linear pair / vertical angle at $P_1$; they characterize the same fact — "the angles cut on the two lines agree".

• Perpendicularity and parallelism: if $t \perp \ell_1$ and $t \perp \ell_2$, then $\angle 1 = \angle 2 = 90^\circ$, and the co-interior sum is exactly $180^\circ$, so $\ell_1 \parallel \ell_2$. This gives the most direct construction route for "drawing a parallel through an external point".

• Bridging forward and backward: this section together with Corresponding/alternate angles ⇔ lines parallel belongs to "neutral geometry" — the $\Rightarrow$ direction holds in hyperbolic and elliptic geometry too; only the converse (Parallel ⇒ co-interior angles supp.) requires the parallel postulate.

Remarks

There is no "figure manipulation" in the proof — flipping $\angle 1$ to $\angle 3$ is just a single linear-pair algebraic step: $180^\circ - \angle 1$. The whole technique is structurally identical to Vertical angles are equal — both arrange for $\angle 1$ to appear in two equations so that subtracting cancels it.

This conclusion is the other half of Euclid I.28: I.28 simultaneously proves "equal corresponding angles ⇒ parallel" and "co-interior angles supplementary ⇒ parallel". We hand the latter to the former, so that the triangle exterior-angle inequality, the "heavy weapon" of neutral geometry, is used only once — at Corresponding/alternate angles ⇔ lines parallel — while the other two criteria are derived from it by purely algebraic shuffling.