Angle at the centre: the angle an arc makes at the centre is twice the angle it makes at the circumference, from any point on the remaining arc. Watch for the reflex-angle version: if the centre angle shown is reflex (e.g. $220°$), the circumference angle is still half of it ($110°$), not half of $360°-220°$.

Angle in a semicircle = $90°$: a diameter is just a "straight line" angle ($180°$) at the centre, so the circumference angle is half of that: $90°$. This is an "enabler" — it hands you a right-angled triangle, which often unlocks Pythagoras or SOHCAHTOA in the next step.

Angles in the same segment are equal: two angles subtended by the same arc, from different points on the same side, must both equal half the same centre angle — so they equal each other. Look for the "bowtie" shape.

A cyclic quadrilateral has all four vertices on the circle (a vertex at the centre breaks this — it's a common trap). Its opposite angles sum to $180°$. If $\angle ABC = w$, then $\angle ADC = 180° - w$.

Radius $\perp$ tangent: a radius meets a tangent at exactly $90°$ at the point of contact — another "enabler" for right-angled-triangle methods.

Two tangents from an external point are equal in length. This also creates an isosceles triangle (so the two base angles are equal), and the line from the external point to the centre bisects both the angle between the tangents and the angle at the centre between the two radii — useful "hidden" facts beyond just the equal lengths.

Example: external point $P$, tangents touching at $A,B$, $\angle APB=40°$: $\triangle PAB$ is isosceles, so $\angle PAB = \angle PBA = \frac{180-40}{2}=70°$.

The angle between a tangent and a chord drawn from the point of contact equals the angle in the alternate segment — the inscribed angle on the other side of that chord. It's the trickiest because its own proof stitches together the semicircle theorem and the radius–tangent rule (draw the diameter from the point of contact), so questions using it are often multi-step by nature.

Treat each given fact as unlocking the next step — write the reason for every line, since "provide reasons" is explicitly assessed.

$BADC$ is a cyclic quadrilateral; a tangent at $A$ makes angle $x$ with chord $AD$ and angle $y$ with... $\angle ABC = w$. Show $x+y=w$.

1. $\angle ADC = 180°-w$ (opposite angles in a cyclic quadrilateral).
2. By the alternate segment theorem, the tangent–chord angles at $A$ equal the inscribed angles in the alternate segments: $\angle DAC=x$ and $\angle DCA=y$.
3. Angles in $\triangle ADC$ sum to $180°$: $(180°-w) + x + y = 180°$.
4. Rearranging: $x+y=w$. ∎

$A$, $B$, $C$ are on a circle, centre $O$. $AB$ is a diameter. $\angle BAC = 30°$. Find $\angle BCA$ and justify each step.

1. $\angle ACB = 90°$ (angle in a semicircle, since $AB$ is a diameter).
2. Angles in $\triangle ABC$ sum to $180°$: $\angle BCA = 180° - 90° - 30° = 60°$.

Whenever a diameter appears in a circle problem, immediately mark the $90°$ it creates — it's almost always the key that opens the rest of the question.