Quoted from Kiselev's geometry: Planimetry, page 18 section 27,
(1) If the sum of several angles that have a common vertex is congruent to a straight angle, then the sum is 180 degrees.
(2) If two angles have a common side and add up to 180 degrees, then their other sides form continuations of each other.
Update: I've cleared up my misunderstanding after reading other text, Serge Lang's Geometry: A High School Course.
Moderator, please delete this post if necessary.
$\endgroup$ 02 Answers
$\begingroup$Suppose that $\angle AOB + \angle BOC = 180^\circ$; we want to prove that $OC$ is a continuation of $AO$.
Construct $OC'$ so that $OC'$ is a continuation of $AO$ and $OC=OC'$. By fact (1), $\angle AOB + \angle BOC' = 180^\circ$, so \begin{align} \angle COC' &= \angle BOC' - \angle BOC \\&= (\angle AOB + \angle BOC') - (\angle AOB + \angle BOC) \\&= 180^\circ - 180^\circ = 0^\circ. \end{align}
From this, it follows that $C=C'$, in some way that depends heavily on the specific axioms you're taking. From my understanding of Kiselev, he'd want us to draw the circle with center $O$ and radius $OC$; if $\angle COC' = 0^\circ$, then the arc $CC'$ has length $0$, so its two endpoints are equal.
$\endgroup$ 0 $\begingroup$As you can see, angles A and B share a common arm(side), and they add upto 180 degrees; since their base forms a straight line at the bottom.
Such angles are said to be forming a linear pair.
Without loss of generality, it can be visualised that their uncommon arms form continuations of each other, making their sum 180 degrees.
