The triangle ABC has circumcentre O, incentre I and orthocentre H.
A, O, I, H and C are concyclic if and only if ∠ABC = 60°.
1. If ∠ABC = 60, then A, O, I, H and C are concyclic.
∠AOC = 120 (angle at centre is twice angle at circumference)
∠BAC + ∠BCA = 120 (angles in a triangle)
But IA and IC bisect these angles, so ∠IAC + ∠ICA = ½∠BAC + ½∠BCA = ½(∠BAC + ∠BCA) = 60
Hence ∠AIC = 120 (angles in a triangle)
Now let the altitudes from A and C meet the opposite sides at A’ and C’ respectively.
∠BC’H = 90 and ∠BA’H = 90, so as angles in quadrilateral BC’HA’ add to 360, ∠A’HC’ = 120
∠AHC = 120 (vertically opposite)
So ∠AOC = ∠AIC = ∠AHC; therefore, AHIOC is cyclic by the converse of the same segment theorem. (The order of these points could be different depending on the triangle.)
2. If A, O, I, H and C are concyclic, then ∠ABC = 60.
Actually, the converse is slightly weaker than this. Only two of O, I and H need to form a cyclic quadrilateral with A and C; the last one of the three will always end up on this circle too. Therefore the ‘converse’ should actually read, ‘Given that A, C and at least two of the points O, I and H are concyclic, then ∠ABC = 60.’
Let ∠ABC = x.
∠AOC = 2x (angle at centre)
∠BAC + ∠BCA = 180 – x (angles in a triangle)
But IA and IC bisect these angles, so ∠IAC + ∠ICA = ½∠BAC + ½∠BCA = ½(∠BAC + ∠BCA) = 90 – ½x
Hence ∠AIC = 90 + ½x (angles in a triangle)
∠BC’H = 90 and ∠BA’H = 90, so as angles in quadrilateral BC’HA’ add to 360, ∠A’HC’ = 180 – x
∠AHC = 180 – x (vertically opposite)
But ∠AOC = ∠AIC = ∠AHC (same segment)
We can see that given any two of the three expressions for these angles in terms of x (2x, 90 + ½x, 180 – x), we can equate them and solve the resulting simultaneous linear equation to obtain x = 60.
(Incidentally, the radii of the circles ABC and AHIOC are equal.)
(Also IH = IO – an angle chase reveals that they subtend equal angles)