H is the orthocentre of an acute-angled triangle ABC. H is reflected in the line BC, and the image is called H′. Where does H′ lie?

Reflecting the orthocentre in a SIDE lands it on the circumcircle — H′ is the second point where the altitude from A meets that circle. The quickest proof is an angle chase: ∠BHC = 180° − ∠A, reflection preserves that angle so ∠BH′C = 180° − ∠A too, and a point seeing BC at the supplement of ∠A lies on the arc BC on the far side, that is on the circumcircle. The trap in the third option is the neighbouring result: reflecting H in the MIDPOINT of BC gives the point diametrically opposite A, a different image point. The nine-point circle carries the MIDPOINT of AH, not the reflection of H, and nothing sends H to the circumcentre.