Cyclic quadrilateral tricks for olympiad geometry

Everybody knows the two facts about a cyclic quadrilateral: opposite angles sum to $180^\circ$, and two angles standing on the same chord from the same side are equal. Knowing them is not what separates a solved geometry problem from a page of labelled angles going nowhere.

What separates them is that in a real olympiad problem the circle is almost never drawn. Four points happen to lie on one, nobody tells you, and the whole solution is noticing. So the useful tricks come in two halves: how to find a cyclic quadrilateral that the problem has hidden, and what to take out of one once you have it.

Five ways to prove four points are concyclic

Each of these is a test. When you suspect four points share a circle, run through the list and see which one your figure can actually supply.

1. Opposite angles are supplementary. If $\angle ABC + \angle CDA = 180^\circ$ in a convex quadrilateral $ABCD$, it is cyclic. This is the converse of the fact proved in the cyclic quadrilateral problem, and it is the one most people try first.
2. Equal angles on a common segment. If $A$ and $D$ are on the same side of line $BC$ and $\angle BAC = \angle BDC$, then $A, B, C, D$ are concyclic. In practice this is the most-used test of all, because angle chasing produces equal pairs far more often than supplementary ones.
3. Two right angles on one segment. If $\angle BXC = \angle BYC = 90^\circ$, both $X$ and $Y$ lie on the circle with diameter $BC$. This is Thales' theorem run backwards, and it is how most hidden circles get found.
4. An exterior angle equals the interior opposite angle. Extend $AB$ past $B$ to $X$. If $\angle CBX = \angle CDA$, the quadrilateral is cyclic. It is test 1 in disguise, but it is the form the angle usually arrives in.
5. A product of lengths through one point. If lines $AB$ and $CD$ meet at $P$ and $PA \cdot PB = PC \cdot PD$ (with the points arranged the same way on each line), then $A, B, C, D$ are concyclic. This is the converse of power of a point, and it is the only test that runs on lengths instead of angles.

A sixth test exists and is worth knowing about rather than reaching for: Ptolemy's inequality $AC \cdot BD \le AB \cdot CD + BC \cdot AD$ becomes an equality exactly when $ABCD$ is cyclic in that order. It is the right tool when a problem is entirely about lengths, and almost never otherwise.

Where the hidden ones sit

Test 3 deserves its own section, because right angles are where hidden circles breed. The richest example is the one every olympiad geometer should be able to draw from memory.

Take an acute triangle $ABC$ with altitudes $AD$, $BE$, $CF$ meeting at the orthocentre $H$. Count the right angles: there are six of them, at $D$, $E$ and $F$, two at each foot. Pair them up and you get six cyclic quadrilaterals without a single circle in the original statement:

• $AEHF$, $BDHF$ and $CDHE$: each has two opposite right angles, so test 1 fires.
• $BCEF$, $CAFD$ and $ABDE$: each has two right angles standing on one side of the triangle, so test 3 fires. For instance, $\angle BEC = \angle BFC = 90^\circ$ puts $E$ and $F$ on the circle with diameter $BC$.

Six circles from three perpendicular lines. Most problems about the orthocentre are solved by picking the right one of those six, and the skill is seeing all six before you start.

What a cyclic quadrilateral hands you back

Once you have a circle, it pays out in three currencies. Know all three, because the question decides which one you want.

• Angles that move. Equal angles on the same chord let you carry an angle from one corner of the figure to another, which is what angle chasing runs on. This is the payout you will use most.
• Products of lengths. Any point on two chords of the circle gives a power-of-a-point identity. You get these for free the moment the circle exists.
• A single length relation. Ptolemy's equality $AC \cdot BD = AB \cdot CD + BC \cdot AD$. Rarely the main event, sometimes the last line.

Watching it work

Here are three standard facts about the orthocentre, each of which falls out in two lines once you pick the right circle from the six.

The altitudes bisect the angles of the orthic triangle. We want $\angle HDF = \angle HDE$. In cyclic $BDHF$, angles on chord $HF$ give $\angle HDF = \angle HBF$, and $\angle HBF = \angle EBA = 90^\circ - \angle A$ from right triangle $ABE$. In cyclic $CDHE$, the same move gives $\angle HDE = \angle HCE = 90^\circ - \angle A$. Equal, so $DA$ bisects angle $EDF$, and $H$ is the incentre of triangle $DEF$. Two circles, no computation.

Three equal products. $ABDE$ is cyclic with chords $AD$ and $BE$ crossing at $H$, so power of a point gives $AH \cdot HD = BH \cdot HE$. Cyclic $BCEF$ gives $BH \cdot HE = CH \cdot HF$. So $AH \cdot HD = BH \cdot HE = CH \cdot HF$: the second payout, collected for free.

The reflected orthocentre lies on the circumcircle. Reflect $H$ in line $BC$ to $H'$. Reflection keeps angles, so $\angle BH'C = \angle BHC$, and from cyclic $AEHF$, $\angle BHC = \angle EHF = 180^\circ - \angle A$. So $\angle BH'C + \angle BAC = 180^\circ$, $H'$ is on the other side of $BC$ from $A$, and test 1 puts $H'$ on the circumcircle of $ABC$.

That last fact is the start of a longer story. The feet $D, E, F$ lie on one more circle, half the size of the circumcircle, which also passes through the side midpoints: the nine-point circle. It is the same orthocentre configuration, one level up.

The tricks, in one place

• Two right angles on a common segment: draw the circle on that segment as diameter.
• An angle chase that keeps producing equal pairs: look for two of them on the same chord, and you have a circle.
• A length product through one point: either a circle already exists, or the converse of power of a point is about to build one.
• A problem entirely about lengths, with four points and no angles: Ptolemy.
• Any altitude in the figure: write down the six orthocentre circles before doing anything else.

How to train it

Draw an acute triangle with its three altitudes and, without looking at the list above, find all six cyclic quadrilaterals and say which test proves each one. Then do the same for a figure with a tangent and a chord. If you can find every circle in a figure in under a minute, the angle chase that follows is mostly bookkeeping.

Lemma teaches angle chasing and power of a point as named techniques, each with its own page and a place in the lesson ladder. The problem archive has the cyclic quadrilateral, Thales, Ptolemy and nine-point circle problems above in full, and the daily problem posts a new one every day in three tiers, no account needed.