Circles: Angles and Cyclic Quadrilaterals

All the inscribed angles on one chord; four equal pairs from the diagonals; whether four points lie on one circle.

A chord $AB$ cuts the circle into two segments. By the last chapter the angle $\angle APB$ at any point $P$ on the chord is always half the central angle, wherever $P$ is on the arc. So keep $A$ and $B$ fixed and slide $P$ along the arc — the angle does not change. This is “angles in the same segment are equal”.

For a point $R$ in the other segment, $\angle ARB=180°-\angle APB$: the angles of the two segments together make half of the full circle ($\frac{360°}{2}=180°$). The converse is true too and is the test for four points to lie on one circle: if $P$ and $Q$ lie on the same side of $AB$ and $\angle APB=\angle AQB$, then $A,B,P,Q$ lie on one circle.

The situation used most: a quadrilateral $ABCD$ with its diagonals $AC$, $BD$. All four sides are chords and each is seen from the same side by the other two points — so there are four pairs of equal angles, as the colours show.

1. Mark the chord of the known angle (the points where its two arms end on the circle).
2. Look at the chord of the asked angle. If it is the same chord and the point is on the same side the angles are equal; on the opposite side, $180°-$.
3. Write the same-coloured angles as equal and continue with the angle sum of a triangle (180°).
4. When the circle is not given, use the converse: equal angles on one side ⟹ the four points are on one circle.

The right angle on a diameter, its converse (the circle round a right triangle) and a chord perpendicular to a diameter.

A diameter $AB$ passes through the centre, so its central angle is the straight angle $180°$. The inscribed angle is half of it: the angle in a semicircle is $90°$. It is the neatest form of the same-segment rule: wherever $P$ goes on the semicircle, $\angle APB$ stays a right angle.

The converse is just as useful: if a triangle has $\angle P=90°$, its hypotenuse $AB$ is a diameter of the circumcircle. So the circumcentre of a right triangle is the midpoint of the hypotenuse, the circumradius is $R=\frac{\text{hypotenuse}}{2}$, and the median to the hypotenuse is half of it. One more result: if a chord $PQ$ perpendicular to $AB$ meets it at $D$, then $P{D}^2=AD\times DB$ (since for crossing chords $PD\cdot DQ=AD\cdot DB$ and $PD=DQ$).

1. If a diameter appears, join its ends to a point of the circle and mark $90°$.
2. If a $90°$ appears and the circle is not given, take the hypotenuse as a diameter (the converse).
3. Find the remaining angles by the angle sum; lengths by Pythagoras or $P{D}^2=AD\cdot DB$.

Four vertices on one circle: opposite angles sum to 180°, the exterior angle, which quadrilaterals are cyclic and the converse test.

A quadrilateral $ABCD$ whose four vertices lie on a circle is a cyclic quadrilateral. $\angle A$ is the inscribed angle standing on the arc $BCD$, so $\angle A=\frac{1}{2}\text{arc }BCD$; $\angle C$ stands on the other arc $BAD$: $\angle C=\frac{1}{2}\text{arc }BAD$. The two arcs together make the whole circle ($360°$), so $\angle A+\angle C=\frac{360°}{2}=180°$. Likewise $\angle B+\angle D=180°$.

Exterior angle: produce $AB$ to $E$; then $\angle ABC+\angle CBE=180°$ (straight line) and $\angle ABC+\angle ADC=180°$ (cyclic), so $\angle CBE=\angle ADC$ — the exterior angle equals the opposite interior angle. Converse: if in a quadrilateral the opposite angles add up to $180°$ (or an exterior angle equals the opposite interior angle) then it is cyclic.

Letters: $\angle A$ and $\angle C$ are opposite interior angles of the quadrilateral; “legs” are the non-parallel sides of a trapezium.

1. Check that the quadrilateral is cyclic (stated, or all four vertices on the circle in the figure).
2. Opposite angles supplementary: $A+C=180°$, $B+D=180°$; exterior = opposite interior.
3. If the angles have $x$ write the equation of an opposite pair; for a ratio use $k$.
4. If diagonals are drawn, add the same-segment angles (previous method).

The link between the sides and diagonals of a cyclic quadrilateral, a point on the arc of an equilateral triangle, and the area of a cyclic quadrilateral.

Ptolemy's theorem: in a cyclic quadrilateral $ABCD$ the product of the diagonals = the sum of the products of the opposite sides. In the figure green = diagonals, indigo = one pair of opposite sides, rose = the other pair.

Why (the idea): take a point $E$ on $AC$ with $\angle ABE=\angle DBC$. By same-segment angles triangles $ABE$ and $DBC$ are similar, so $AE\cdot BD=AB\cdot CD$; and triangles $ABD$ and $EBC$ are similar, so $EC\cdot BD=AD\cdot BC$. Adding, $(AE+EC)\cdot BD=AC\cdot BD=AB\cdot CD+AD\cdot BC$.

The second form gives the ratio of the diagonals (each diagonal from the products of the sides at its ends). The converse holds too: a quadrilateral with $AC\cdot BD=AB\cdot CD+AD\cdot BC$ is cyclic. Special cases: a rectangle has equal diagonals $d$ and $d^2=a^2+b^2$ (Pythagoras); an isosceles trapezium (bases $a,b$, leg $c$) has diagonal $d^2=ab+c^2$.

1. Write the vertices in order and mark the two pairs of opposite sides.
2. Product of the diagonals = product of one pair + product of the other pair.
3. Solve for the unknown (a side or a diagonal); for the area find $s$ and use Brahmagupta.

The angle between two chords crossing inside or two secants meeting outside, from the arcs; also in terms of central angles.

So far the vertex of the angle was either at the centre or on the circle. When the vertex is inside (two chords crossing) or outside (two secants meeting), the angle comes from the arcs. Inside: if chords $AB$ and $CD$ cross at $P$, then $\angle APC=\frac{1}{2}(\text{arc }AC+\text{arc }BD)$ — half the sum of the two arcs that the angle and its vertically opposite angle face.

Why? $\angle APC$ is an exterior angle of triangle $PBC$, so $\angle APC=\angle ABC+\angle DCB$ (the two remote interior angles). But $\angle ABC$ is an inscribed angle on the arc $AC$ ($=\frac{1}{2}\text{arc }AC$) and $\angle DCB$ on the arc $DB$ ($=\frac{1}{2}\text{arc }DB$). Adding gives the formula. Outside: for secants $PAB$, $PCD$ ($A,C$ the near points) $\angle P=\frac{1}{2}(\text{far arc }BD-\text{near arc }AC)$ — the exterior angle $\angle BAD=\angle P+\angle ADP$ gives $\angle P=\angle BAD-\angle ADC$.

1. Mark the ends of both arms of the angle and see which arc lies inside it.
2. Vertex inside: add the two arcs (the angle's and its vertically opposite one) and halve.
3. Vertex outside: far arc − near arc, then halve.
4. If an arc is unknown, turn the formula round; if central angles are given they are the same numbers as the arcs.

The strategy for angle hunts when the centre is in the figure: which tools to use, in which order.

Angle hunts with a centre $O$ need no new theorem — they are the art of using the old tools in the right order. There are four tools: (1) two radii = an isosceles triangle (equal base angles); (2) centre angle = 2 × circle angle; (3) the perpendicular from the centre to a chord (midpoint, $90°$); (4) the angle sum of a triangle / quadrilateral ($180°$, $360°$).

Letters: $\angle OAB$ is the angle between the radius $OA$ and the chord $AB$; $\angle AOB$ the central angle; $M,N$ the midpoints of the chords.

1. Draw the figure: $O$, all the given chords, the lines joining the midpoints to $O$.
2. Start from the easiest angle (a base angle of an isosceles triangle, or $90°$ from a straight line/perpendicular).
3. Feed every new angle into the next theorem: ×2 or ÷2, or a triangle sum.
4. Stop when the asked angle appears, and verify by another route.

The bisector of angle $A$ meets the circumcircle again at $E$: $E$ is an arc midpoint, three similar pairs and their products.

Take triangle $ABC$ and its circumcircle. The bisector of $\angle A$ meets $BC$ at $D$ and the circle again at $E$. This figure appears again and again in exams because it carries four useful facts at once.

1. $E$ is the midpoint of the arc $BC$: $\angle BAE=\angle EAC$ (bisector) are equal inscribed angles, so their arcs $BE$ and $EC$ are equal, i.e. $EB=EC$.
2. $△ABD\sim △AEC$: $\angle BAD=\angle EAC=\alpha$ and $\angle ABD=\angle ABC=\angle AEC$ (chord $AC$, same segment). So $\frac{AB}{AE}=\frac{AD}{AC}$, i.e. $AB\cdot AC=AD\cdot AE$.
3. $△EBD\sim △EAB$: $\angle E$ is common and $\angle EBD=\angle EBC=\angle EAC=\angle BAE$. So $E{B}^2=ED\cdot EA$.
4. Crossing chords $AE$ and $BC$: $AD\cdot DE=BD\cdot DC$. Putting $AE=AD+DE$ in (2): $AB\cdot AC=A{D}^2+AD\cdot DE=\mathbf{A}{\mathbf{D}}^2+\mathbf{BD}\cdot \mathbf{DC}$.

One more identity (equal chords): if $CA=CB$ and a line through $C$ meets $AB$ (produced beyond $A$) at $P$ and the circle again at $T$, then $CT\cdot CP=C{A}^2$. Reason: in triangles $CTA$ and $CAP$ the angle $C$ is common and $\angle CTA=\angle CBA=\angle CAB$ (same segment and $CA=CB$), so they are similar.