Lesson · Concyclicity
Proving four points concyclic
"Prove that these four points lie on one circle" looks like a demand for a construction. It is not. Every proof of concyclicity is one of four short angle or distance checks, and choosing the right one is the whole task.
The four tests
| Test | Conclusion |
|---|---|
| \(P\), \(Q\) on the same side of \(AB\) and \(\angle APB = \angle AQB\) | \(A\), \(B\), \(P\), \(Q\) concyclic |
| \(APBQ\) convex with \(\angle APB + \angle AQB = 180^\circ\) | \(A\), \(B\), \(P\), \(Q\) concyclic |
| \(\angle APB = \angle AQB = 90^\circ\) | all four on the circle with diameter \(AB\) |
| \(MA = MB = MP = MQ\) for some point \(M\) | circle with centre \(M\) |
The first two are the converse of the inscribed angle theorem; the third is its right-angled special case; the fourth is the definition, and is the one to use when the candidate centre is visible in the figure (a midpoint, or the meeting point of two perpendicular bisectors).
The Thales-circle habit
Principle
Whenever two right angles in a figure stand over the same segment, stop and name the circle with that segment as diameter. Altitudes, perpendiculars dropped from a point, and rectangles hand out right angles for free, so this circle is usually already in the picture.
Example
Let \(BB_0\) and \(CC_0\) be altitudes of triangle \(ABC\). Both \(\angle BB_0C\) and \(\angle BC_0C\) are right angles over the segment \(BC\), so \(B\), \(C\), \(B_0\), \(C_0\) lie on the circle with diameter \(BC\). Reading the circle backwards, \(BCB_0C_0\) is a cyclic quadrilateral, hence \(\angle AC_0B_0 = \angle ACB\): the two altitude feet cut off a triangle similar to \(ABC\).
Choosing the segment
With four given points there are three ways to split them into "the chord" and "the two viewers", and only one of them is usually provable. Scan the figure for the pair whose viewing angles you can actually compute - typically a pair joined by a segment that already carries right angles, equal angles, or a known midpoint.
Warning
If your four points are not visibly in convex position, the two tests behave differently: equal angles need the viewers on the same side of the chord, supplementary angles need them on opposite sides. State which case you are in, or handle both.
Warm-up
Example
In a convex quadrilateral \(ABCD\) the diagonals meet inside, and \(\angle ADB = \angle ACB\). Prove that \(\angle DAC = \angle DBC\).
The hypothesis is exactly one of the four tests
\(C\) and \(D\) lie on the same side of the line \(AB\) and see \(AB\) under equal angles.
Draw the circle
So \(A\), \(B\), \(C\), \(D\) lie on one circle. Now the conclusion is a forward use of the same theorem.
Which chord do the new angles stand on?
Both \(\angle DAC\) and \(\angle DBC\) are inscribed angles over the chord \(DC\), with \(A\) and \(B\) on the same arc - so they are equal. Once a circle is on the table, every angle in the figure can be read off it.