**ABC is a triangle with circumcentre O, orthocentre H and circumradius R. P is a point on the perimeter of the triangle. Then OP + PH ≥ R.**

**Proof**

P lies on one of the sides of the triangle (let’s say this includes both the edge’s vertices), so assume WLOG that this side is BC as above.

Let the reflection of H across BC be H’. It is well known that H’ lies on the circumcircle of triangle ABC.

Then OP + PH = OP + PH’. The length of the path OPH’ is at least the straight-line distance from O to H’, which is R, with equality if and only if O, P and H’ are collinear (ie if ∠OPB = ∠HPC).

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