Vertical angles — Vertical angles are equal · Principia

Depends on: Linear pair sums to 180°.

Statement

Let two distinct lines $\ell_1$ , $\ell_2$ intersect at a point $O$ . Four angles are formed at this intersection; any two angles that share no edge are called a pair of vertical angles, and there are two such pairs — their sides are given by the two opposite rays from $O$ along $\ell_1$ and $\ell_2$ respectively.

Take points $A$ , $B$ on $\ell_1$ on opposite sides of $O$ , and points $C$ , $D$ on $\ell_2$ on opposite sides of $O$ . Then the two pairs of vertical angles are equal:

\angle AOC = \angle BOD,\qquad \angle AOD = \angle BOC.

$Two lines intersecting at O: \angle AOC = \angle BOD = \alpha, \angle AOD = \angle BOC = \beta.$

Proof

We only use Linear pair sums to 180°. Let the same angle $\angle AOD$ appear in two linear-pair equations at once; subtraction cancels it.

Application 1: at the intersection point $O$ , apply Linear pair sums to 180° to line $\ell_2$ (which contains $C$ and $D$ , with the opposite rays $OC$ , $OD$ from $O$ ) and ray $OA$ ( $OA\subset\ell_1$ , and since $\ell_1\neq\ell_2$ , $OA$ is not on $\ell_2$ , satisfying the "non-collinear" hypothesis of Linear pair sums to 180°):

\angle AOC + \angle AOD = 180^\circ. \tag{1}

$Step 1: cutting off the linear-pair \angle AOC and \angle AOD on line \ell_2 with ray OA, linear-pair gives (1).$

Application 2: at $O$ , apply Linear pair sums to 180° to line $\ell_1$ (opposite rays $OA$ , $OB$ ) and ray $OD$ ( $OD\subset\ell_2$ , $OD$ not on $\ell_1$ ):

\angle AOD + \angle BOD = 180^\circ. \tag{2}

$(1) - (2)$ cancels the $\angle AOD$ term:

\angle AOC - \angle BOD = 0,\quad\text{i.e.}\quad \angle AOC = \angle BOD.

$Step 2: (1) and (2) share \angle AOD; subtraction cancels it and gives the first vertical-angle equality \angle AOC = \angle BOD.$

The second pair of vertical-angle equalities follows from one more application of Linear pair sums to 180°.

Application 3: at $O$ , apply Linear pair sums to 180° to line $\ell_1$ and ray $OC$ :

\angle AOC + \angle BOC = 180^\circ. \tag{3}

$(1) - (3)$ cancels the $\angle AOC$ term:

\angle AOD - \angle BOC = 0,\quad\text{i.e.}\quad \angle AOD = \angle BOC. \qquad\blacksquare

$Step 3: (1) and (3) share \angle AOC; subtraction cancels it and gives the second vertical-angle equality \angle AOD = \angle BOC.$

Immediate consequences

The "scissors" four angles. The four angles formed by two intersecting lines pair up into two pairs of equal vertical angles and two pairs of supplementary linear pairs. Overall, the four angles sum to exactly $360^\circ$ .

$Scissors diagram: two lines meeting at O, the two pairs of vertical angles are \alpha and \beta; \alpha + \beta = 180^\circ, 2\alpha + 2\beta = 360^\circ.$

Transitivity of perpendicularity. If $\ell_1\perp\ell_2$ , then all four angles are $90^\circ$ (one pair of equal vertical angles + one pair of supplementary linear pairs equal to $180^\circ$ gives this immediately).
"Algebraization of angles." Vertical angles are equal is the first place where, on a figure of two crossing lines, we can shuffle angles around as if manipulating algebraic expressions — a technique that recurs throughout the proofs of "corresponding / alternate interior angles ⇔ parallel."

Remarks

The relationship between Linear pair sums to 180° and vertical angles are equal is one of the cleanest "reuse" examples in geometry: Linear pair sums to 180° carves "a straight line = $180^\circ$ " into a theorem, and vertical angles are equal applies that theorem to two lines simultaneously, so that the term $\angle AOD$ appears in both equations and cancels by subtraction. The whole proof contains no "figure manipulation" — no translation, no rotation, just two substitutions and one subtraction.

This conclusion is I.15 in Euclid's Elements, where Euclid uses the same two-linear-pairs strategy; the only difference is that in the Elements the very statement "the angles are equal" depends on an implicit moving-figure argument, while here we land directly on axiom III, making the entire proof fully algebraic.