The lines connecting the circumcentre of a triangle ABC and its vertices are each perpendicular to a side of the orthic triangle of ABC.
I am going to prove this for the case where the circumcentre lies within the orthic triangle. I think this proof can be used for all other configurations as well (see below) – all you have to do is extend the lines involved.
The acute case:
Variant configurations: A) ABC is acute, but the circumcentre is outside orthic triangle
B) Original triangle is obtuse – orthic triangle isn’t even contained within the circumcircle.
Proof for acute case:
Let D, E and F be the feet of the perpendiculars dropped from A to BC, B to AC and C to AB respectively. Let O be ABC’s circumcentre, and H its orthocentre. Let OC cut ED at P.
Let angle BAC be . Angle AFC = 90, so by angles in a triangle, angle FCA = 90 –
Also, recall that DCEH is cyclic (since angle HDC + angle HEC = 90 + 90 = 180, it is cyclic by converse of opposite angles in a cyclic quadrilateral)
So, using angles in same segment, angle HCE = angle HDE = 90 – .
Since HD and BC are perpendicular, angle PDC = .
Also, BOC = , by angle at centre is twice angle at circumference
BO = OC (both radii of circumcircle), so BOC is iscosceles; hence by angle in a triangle, angle OCD = 90 –
By angles in a triangle therefore, angle DPC = 180 – – (90 – ) = 90
We apply this to the other sides of the triangle and the result is proven.