Only basic axioms — never the theorem you're trying to prove (that would be circular reasoning). Your toolkit: angles in a triangle sum to $180°$; angles on a line sum to $180°$; angles round a point sum to $360°$; base angles of an isosceles triangle are equal; exterior angle = sum of opposite interior angles; and conditions for congruence (RHS, SSS, SAS, ASA).

Draw the radius. Any two radii drawn to points on the circumference form an isosceles triangle (since all radii are equal), and the equal base angles become your algebraic foothold — label them $x$ and $y$ and work from there.

This is the master proof — nearly every other angle theorem follows from it.

Construction: Join $A,B$ (circumference) to centre $O$ and to a point $P$ on the circumference; extend $PO$ to $Q$.

Let $\angle OPA=x,\ \angle OPB=y$, so $\angle APB=x+y$.

Since $OA=OP$ (radii), $\triangle APO$ is isosceles: $\angle OAP=\angle OPA=x$. The exterior angle rule gives $\angle AOQ = x+x = 2x$. By the same logic on $\triangle BPO$: $\angle BOQ=2y$.

QED. (Three configurations exist depending on whether $O$ lies inside, on, or outside $\angle APB$ — the reflex-angle version of this proof is what makes the cyclic quadrilateral proof below possible.)

Questions sometimes explicitly forbid using the master proof above, forcing this "ab initio" version.

Construction: $AB$ is a diameter, centre $O$; $C$ is any point on the circumference. Draw radius $OC$. Let $\angle OAC=x,\ \angle OBC=y$.

Since $OA=OC$ and $OB=OC$ (radii), both $\triangle AOC$ and $\triangle BOC$ are isosceles, so $\angle OCA=x$ and $\angle OCB=y$.

Angles in $\triangle ABC$ sum to $180°$: $x+y+(x+y)=180° \implies 2(x+y)=180° \implies x+y=90°$.

Since $\angle ACB=x+y$, $\angle ACB=90°$. QED.

Let $\angle AOB=x$ (angle at centre). By the master proof, $\angle AOB=2\times\angle ACB$ and $\angle AOB=2\times\angle ADB$ (both $C,D$ subtend the same arc $AB$). So $2\angle ACB=2\angle ADB \implies \angle ACB=\angle ADB$. QED.

Join centre $O$ to opposite vertices $B,D$ of cyclic quadrilateral $ABCD$. The reflex $\angle BOD = 2\angle DAB$ (angle at centre, major arc) and the minor $\angle BOD = 2\angle BCD$ (angle at centre, minor arc). Since these two angles at $O$ together complete a full turn: $2\angle DAB+2\angle BCD=360° \implies \angle DAB+\angle BCD=180°$. QED.

This one uses proof by contradiction, not algebra. Any point $Q\ne P$ on the tangent line must lie outside the circle (since the tangent touches at only one point $P$), so $OQ>OP$ for every such $Q$. This means $OP$ is the shortest distance from $O$ to the tangent line — and the shortest distance from a point to a line is always the perpendicular. So $OP\perp$ tangent. QED.

Join external point $P$ to centre $O$, and to the two points of contact $A,B$. In $\triangle PAO$ and $\triangle PBO$: both have a right angle (radius⊥tangent), share hypotenuse $OP$, and have $OA=OB$ (radii) — RHS congruence. Congruent triangles have equal corresponding sides, so $PA=PB$. QED.

The hardest proof on the specification — a multi-step chain. Tangent at $A$, chord $AC$, point $B$ in the alternate segment; let $\angle TAC=x$. Draw the diameter $AD$ through $A$, join $D$ to $C$.

1. $\angle TAD=90°$ (tangent⊥radius/diameter)
2. $\angle DAC = 90°-x$ (angles making up $\angle TAD$)
3. $\angle ACD=90°$ (angle in semicircle, subtended by diameter $AD$)
4. Angles in $\triangle ADC$ sum to $180°$: $\angle ADC+90°+(90°-x)=180° \implies \angle ADC=x$
5. $\angle ABC=\angle ADC$ (angles in the same segment, both subtend arc $AC$)
6. So $\angle ABC=x=\angle TAC$. QED.

$A,B,C,D$ are points on a circle. $PCQ$ is a tangent at $C$. $AB=CB$. $\angle BCQ=x°$. Prove $\angle CDA=2x°$.

1. $\angle BAC=x°$ — Alternate Segment Theorem.
2. $\angle BCA=\angle BAC=x°$ — base angles of isosceles $\triangle ABC$ (since $AB=CB$).
3. $\angle ABC=180°-2x°$ — angles in a triangle.
4. $\angle CDA+\angle ABC=180°$ — opposite angles of cyclic quadrilateral $ABCD$.
5. $\angle CDA = 180°-(180°-2x°) = 2x°$. QED.