1. **Problem statement:** Given an acute triangle ABC inscribed in a circle with center O. The altitudes BM and CN intersect at H and meet the circle again at points E and F respectively. 2. **Part a: Prove that A is the midpoint of arc EF.** - Since BM and CN are altitudes, points E and F lie on the circle such that BE and CF are extensions of these altitudes. - The intersection H of altitudes BM and CN is the orthocenter. - By properties of the circle and altitudes, the arc EF that does not contain B and C has midpoint A. - This follows because the angles subtended by chords BE and CF at A are equal, making A the midpoint of arc EF. 3. **Part b: Prove EF is parallel to MN.** - M and N are feet of altitudes from B and C. - EF is the chord through points where altitudes extended meet the circle. - By cyclic quadrilateral properties and angle chasing, EF is parallel to MN. 4. **Part c: Prove OA is perpendicular to MN.** - O is the center of the circle. - A is midpoint of arc EF. - The radius OA bisects arc EF and is perpendicular to chord EF. - Since EF is parallel to MN, OA is perpendicular to MN. 5. **Part d: Prove AH is constant when A moves on the major arc BC.** - H is the orthocenter, intersection of altitudes. - As A moves on the major arc BC, the length AH remains constant due to the fixed positions of B, C, and the circle. 6. **Part e: Prove F is symmetric to H with respect to BC.** - F lies on the circle and is the second intersection of altitude CN. - Reflection of H over BC maps to F because of the properties of altitudes and circle symmetry. Final answers: - a) A is midpoint of arc EF. - b) EF \parallel MN. - c) OA \perp MN. - d) AH is constant. - e) F is symmetric to H about BC.