1. **State the problem:** We have a circle with center O and points A, B, C, D on the circumference. AB is a diameter, |BC| = |CD|, and angle AÔD = 64°. We need to find all angles of quadrilateral ABCD. 2. **Analyze given information:** - Since AB is a diameter, angle AÔB = 180°. - Angle AÔD = 64° is given. - |BC| = |CD| means triangle BCD is isosceles with BC = CD. 3. **Find angle BÔD:** Since AÔB = 180°, and AÔD = 64°, then $$B\hat{O}D = 180^\circ - 64^\circ = 116^\circ.$$ 4. **Find angles at center for triangle BCD:** Since BC = CD, triangle BCD is isosceles with base BD. The central angles BÔC and CÔD correspond to arcs BC and CD, which are equal. So, $$B\hat{O}C = C\hat{O}D = x.$$ We know $$B\hat{O}C + C\hat{O}D + B\hat{O}D = 360^\circ,$$ so $$x + x + 116^\circ = 360^\circ,$$ $$2x = 244^\circ,$$ $$x = 122^\circ.$$ 5. **Find angles at circumference:** - Angle at circumference subtended by diameter AB is 90°, so $$A\hat{B}C = 90^\circ.$$ - Angles subtended by equal arcs BC and CD at circumference are equal: $$B\hat{C}D = C\hat{D}A = \frac{1}{2} B\hat{O}C = \frac{122^\circ}{2} = 61^\circ.$$ 6. **Find angle at A:** Sum of angles in quadrilateral ABCD is 360°, so $$A\hat{B}C + B\hat{C}D + C\hat{D}A + D\hat{A}B = 360^\circ,$$ $$90^\circ + 61^\circ + 61^\circ + D\hat{A}B = 360^\circ,$$ $$D\hat{A}B = 360^\circ - 212^\circ = 148^\circ.$$ **Final angles of quadrilateral ABCD:** $$A\hat{B}C = 90^\circ,$$ $$B\hat{C}D = 61^\circ,$$ $$C\hat{D}A = 61^\circ,$$ $$D\hat{A}B = 148^\circ.$$