1. **Problem statement:** Given a circle with points A, B, C, D, E on the circumference and center O, with EOB a straight line, $\angle AÔB = 105^\circ$ and $\angle BÊC = 20^\circ$. Find the sizes of: (a) $\angle A\hat{B}C$ (b) $\angle A\hat{D}C$ (c) $\angle B\hat{C}D$ (d) $\angle D\hat{A}B$ 2. **Key formulas and rules:** - The angle at the center $\angle AÔB$ subtends the arc $AB$. - The angle at the circumference subtending the same arc is half the central angle: $\angle A\hat{B}C = \frac{1}{2} \angle AÔB$. - Angles on a straight line sum to $180^\circ$. - Opposite angles in cyclic quadrilaterals sum to $180^\circ$. 3. **Step (a): Find $\angle A\hat{B}C$** - $\angle AÔB = 105^\circ$ is the central angle subtending arc $AB$. - The angle at the circumference subtending the same arc $AB$ is half of $105^\circ$: $$\angle A\hat{B}C = \frac{1}{2} \times 105^\circ = 52.5^\circ$$ 4. **Step (b): Find $\angle A\hat{D}C$** - $\angle BÊC = 20^\circ$ is an angle at point E on the circumference. - Since EOB is a straight line, $\angle BÊC$ subtends arc $BC$. - The angle $\angle A\hat{D}C$ subtends the same arc $BC$ but from point D. - By the circle theorem, angles subtending the same arc are equal: $$\angle A\hat{D}C = \angle BÊC = 20^\circ$$ 5. **Step (c): Find $\angle B\hat{C}D$** - Consider quadrilateral ABCD inscribed in the circle. - Opposite angles in a cyclic quadrilateral sum to $180^\circ$. - $\angle A\hat{B}C = 52.5^\circ$ (from step a) is opposite to $\angle B\hat{C}D$. So, $$\angle B\hat{C}D = 180^\circ - 52.5^\circ = 127.5^\circ$$ 6. **Step (d): Find $\angle D\hat{A}B$** - Consider triangle ABD. - $\angle D\hat{A}B$ subtends arc $DB$. - Arc $DB$ is the remaining arc after arcs $AB$ and $BC$. - Central angle $\angle EÔB = 180^\circ$ since EOB is a straight line. - Arc $AB$ corresponds to $105^\circ$ (central angle), arc $BC$ corresponds to $2 \times 20^\circ = 40^\circ$ (since $\angle BÊC = 20^\circ$ is half the arc). - So arc $DB = 360^\circ - (105^\circ + 40^\circ) = 215^\circ$. - Angle at circumference subtending arc $DB$ is half the arc: $$\angle D\hat{A}B = \frac{1}{2} \times 215^\circ = 107.5^\circ$$ **Final answers:** (a) $\angle A\hat{B}C = 52.5^\circ$ (b) $\angle A\hat{D}C = 20^\circ$ (c) $\angle B\hat{C}D = 127.5^\circ$ (d) $\angle D\hat{A}B = 107.5^\circ$