PRINCIPIA · THEOREM

Linear pair sums to 180°

Dependencies: Protractor axiom.

Statement

Let OO be a point on the line \ell, and let CC, DD be points on \ell on opposite sides of OO; take a further point EE not on \ell. Then the two angles in the linear pair sum to

COE+EOD=180.\angle COE + \angle EOD = 180^\circ.

Here a "linear pair" means two angles sharing the vertex OO and one ray OEOE, while their other rays OCOC, ODOD lie on the same line in opposite directions.

Linear pair: \angle COE + \angle EOD = 180^\circ

Proof

The proof uses only two direct consequences of Protractor axiom (axiom III).

Step 1: The angle between opposite rays is 180180^\circ.

Axiom III sets up a one-to-one correspondence between the rays from OO and the real numbers in [0,2π)[0,\,2\pi); the angle between two rays equals the difference of the corresponding numbers (taking the representative in [0,π][0,\,\pi]).

For the two opposite rays OCOC, ODOD from OO on line \ell, the difference of the numbers assigned by axiom III is exactly π\pi. Hence

COD=π=180.\angle COD = \pi = 180^\circ.

Step 1: opposite rays OC, OD at O span the 180^\circ straight angle

Step 2: Angle additivity splits COD\angle COD.

The ray OEOE is not collinear with \ell (since EE is not on \ell), so by the plane separation implied by the continuity of axiom III, the line \ell divides the plane into two disjoint open half-planes, and EE must lie in one of them. The two cases are symmetric, so we may as well take EE to lie in the "upper" half-plane.

View COD\angle COD as the 180180^\circ straight angle "pointing into the upper half" — its interior is exactly the upper half-plane minus the point OO. The ray OEOE leaves OO pointing toward EE, so it lies entirely inside this straight angle. By the angle additivity of axiom III,

COE+EOD=COD.\angle COE + \angle EOD = \angle COD.

Step 3: Combine.

Substituting step 1 into step 2:

COE+EOD=180.\angle COE + \angle EOD = 180^\circ. \qquad\blacksquare

Steps 2 + 3: ray OE splits the 180^\circ straight angle \angle COD into the linear pair, summing to \angle COE + \angle EOD = 180^\circ

Immediate consequences

  • Entry point to triangle exterior angles: in ABC\triangle ABC, extending the side BCBC past CC to a point XX, the angles ACB\angle ACB and ACX\angle ACX share the vertex CC and the ray CACA, while their outer rays CBCB, CXCX point in opposite directions along line BCBC. By the linear pair = 180° identity:
ACB+ACX=180.\angle ACB + \angle ACX = 180^\circ.

This is the entry point for Exterior angle equals the sum of two non-adjacent interior angles.

  • Quantitative definition of "perpendicular": if at some point a pair of linear-pair angles formed by two lines are equal, then each angle equals 1802=90\dfrac{180^\circ}{2}=90^\circ. This is precisely the standard definition of "\ell\perp\ell'".

  • Entry point to the vertical-angle equality: applying linear pair = 180° to two pairs of linear-pair angles and subtracting immediately yields Vertical angles are equal — see the proof of the next theorem.

Remarks

Linear pair = 180° looks "obvious", but it actually formally welds together two facts:

  • Axiom II tells us "two points determine a line" — but the line itself carries no information about "spanning 180180^\circ";
  • Axiom III assigns each ray a number in [0,2π)[0,\,2\pi) and stipulates that opposite rays correspond to numbers differing by π\pi.

It is this convention of axiom III that makes "line = 180180^\circ" hold. Linear pair = 180° packages this convention together with angle additivity into a theorem that can be cited repeatedly; from this point on, every argument involving angle addition or subtraction (Vertical angles are equal, Triangle interior angles sum to 180°, Exterior angle equals the sum of two non-adjacent interior angles, Central angle is twice the inscribed angle on the same arc, …) starts from linear pair = 180°: cut a line at some point, and add or subtract the two resulting angles.

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