O is the circumcentre and H the orthocentre of triangle ABC. The circumradius is 10 cm and the perpendicular distance from O to the side BC measures 6 cm. What is the length of AH?

A standard Euler-line result says the distance from a vertex to the orthocentre is twice the distance from the circumcentre to the opposite side: AH = $2\times OM$, where $M$ is the midpoint of BC. So AH = $2\times 6=12$ cm. (The reason is that $OM$ joins the midpoints of BC and of the segment from B to... more simply, the medial triangle is the half-size copy of ABC and $O$ is its orthocentre, so the two distances scale by 2.) The trap 8 cm is the half-chord √$(10^2-6^2)$, which gives BC = 16 cm and is not asked for; 6 cm copies $OM$ straight across; and 20 cm is the diameter $2R$, which would be AH only if ∠A were vanishingly small.