Concyclic Points

In one sentence

A circle comes with lots of “free” equalities: inscribed angles on the same arc are equal, opposite angles of a cyclic quadrilateral are supplementary, and so on. But problems often don't draw the circle. The concyclic-points method is: first use angles or lengths to prove that four points lie on one circle, then use every property of that invisible circle.

So you need every test both ways round:

Test (proves concyclic)	Property (usable once concyclic)
$C$ , $D$ on the same side of $A B$ and $∠ A C B = ∠ A D B$	Inscribed angles on the same arc are equal
A pair of opposite angles of the quadrilateral is supplementary	Opposite angles of a cyclic quadrilateral are supplementary
An exterior angle equals the opposite interior angle	Exterior angle = opposite interior angle
The diagonals meet at $P$ and $P A \cdot P C = P B \cdot P D$	The intersecting chords theorem (next stop)

All four tests say the same thing

Step 1 / 6

D inside the circle
The dashed curve is the circle through $A$ , $B$ , $C$ . While $D$ is inside it, the angle $∠ A D B$ that $A B$ subtends at $D$ is bigger than the inscribed angle $∠ A C B$ .

$∠ A D B > ∠ A C B$
D outside the circle
The moment they are equal
Opposite angles are supplementary
Exterior angle = opposite interior angle
Equal products

Predict first

Which of these quadrilaterals always has a circumcircle (all four vertices concyclic)?

A rhombusA general parallelogramAn isosceles trapezoidA right trapezoid

Choose an answer first; then you can see the result.

Stop 1 · Two right angles on the same segment

The most common signal for concyclic points is a right angle. An inscribed angle on a diameter is $9 0^{\circ}$ ; conversely, if $∠ A C B = ∠ A D B = 9 0^{\circ}$ , then $C$ and $D$ both lie on the circle with diameter $A B$ .

Example: in $△ A B C$ , $B E$ and $C F$ are two altitudes. $∠ B E C = ∠ B F C = 9 0^{\circ}$ , and $E$ and $F$ are on the same side of $B C$ , so $B$ , $C$ , $E$ , $F$ are concyclic, and the center of the circle is the midpoint of $B C$ .

Here

A

and

B

are the ends of a diameter. Drag

D

onto the circle: what does

∠ A D B

read?

Stop 2 · Two right angles facing each other

If two right angles are a pair of opposite angles of a quadrilateral, they add up to $18 0^{\circ}$ , so again the vertices are concyclic.

Example: the two altitudes above meet at $H$ . In quadrilateral $A F H E$ , $∠ A F H = ∠ A E H = 9 0^{\circ}$ , so $A$ , $F$ , $H$ , $E$ are concyclic, with diameter $A H$ . Hence $∠ F H E = 18 0^{\circ} - ∠ A$ , and the angle at the orthocenter falls right out.

Stop 3 · One problem, two solutions

One problem, several solutions

In acute $△ A B C$ , $B E ⊥ A C$ at $E$ and $C F ⊥ A B$ at $F$ . Prove that $∠ A F E = ∠ A C B$ .

$∠ B E C = ∠ B F C = 9 0^{\circ}$ and $E$ , $F$ are on the same side of $B C$ , so $B$ , $C$ , $E$ , $F$ are concyclic.

$∠ A F E$ is an exterior angle of the cyclic quadrilateral $B C E F$ , so it equals the opposite interior angle $∠ E C B$ , which is $∠ A C B$ .

CostTwo lines, once you spot “two right angles on BC”.

Similar triangles

$△ A B E$ and $△ A C F$ are both right triangles sharing $∠ A$ , so $△ A B E \sim △ A C F$ , which gives $\frac{A E}{A F} = \frac{A B}{A C}$ .

Rewrite it as $\frac{A E}{A B} = \frac{A F}{A C}$ . Together with the common angle $∠ A$ this gives $△ A E F \sim △ A B C$ , so $∠ A F E = ∠ A C B$ .

CostNo circle, but two similarity proofs.

The two solutions are really the same: the proportion $A E \cdot A C = A F \cdot A B$ in the second one is exactly the “power of a point” of the next stop.

Practice

Exercise 1Basic

Quadrilateral $A B C D$ is inscribed in a circle, and $∠ A = 7 5^{\circ}$ . Find $∠ C$ in degrees.

Exercise 2Basic

In cyclic quadrilateral $A B C D$ , $∠ A : ∠ B : ∠ C = 2 : 3 : 7$ . Find $∠ D$ in degrees.

Exercise 3Advanced

In acute $△ A B C$ , $∠ A = 5 0^{\circ}$ , and the altitudes $B E$ and $C F$ meet at $H$ . Find $∠ B H C$ in degrees.

Exercise 4Challenge

In acute $△ A B C$ the altitudes $A D$ and $B E$ meet at $H$ . Extend $A D$ to meet the circumcircle of $△ A B C$ at $K$ . Prove that $H D = D K$ .

Concyclic Points

In one sentence

All four tests say the same thing

Stop 1 · Two right angles on the same segment

Stop 2 · Two right angles facing each other

Stop 3 · One problem, two solutions

Concyclic points

Similar triangles

Practice

Mistake Museum · Exhibits from this stop

Equal angles, so concyclic — without checking they are on the same side

Adding two adjacent angles to get 180°