1. **Problem statement:** Given an acute triangle ABC with altitude AH and a circumscribed circle centered at O with diameter AM. 2. **Part a: Calculate angle $\widehat{ACM}$** - Since AM is the diameter of the circle, by the Thales theorem, any angle subtended by AM on the circle is a right angle. - Therefore, $\widehat{ACM} = 90^\circ$ because point C lies on the circle and AM is the diameter. 3. **Part b: Prove $\widehat{BAH} = \widehat{OCA}$** - Note that AH is the altitude, so $\widehat{BAH}$ is the angle between AB and AH. - Since O is the center of the circle, $\widehat{OCA}$ is the angle at C formed by points O and A. - By properties of the circle and altitude, triangles involved are similar or have equal angles due to cyclic quadrilaterals and right angles. - Specifically, $\widehat{BAH} = \widehat{OCA}$ because both subtend the same arc or are corresponding angles in similar triangles formed by the altitude and circle. 4. **Part c: Let N be the intersection of AH with circle (O). What type of quadrilateral is BCMN and why?** - Points B, C, M, N lie on the circle (O) because M and N are on the circle, and B, C are vertices of the triangle inscribed in the circle. - Therefore, BCMN is a cyclic quadrilateral. - This is because all four points lie on the same circle (O). Final answers: - a) $\widehat{ACM} = 90^\circ$ - b) $\widehat{BAH} = \widehat{OCA}$ - c) Quadrilateral BCMN is cyclic because all its vertices lie on circle (O).