1. **Problem statement:** Given a circle with center O and points A, B, C, D, E on the circumference, where DE = DC, reflex angle EOC = 240°, and angle EAB = 100°. We need to find angles DCE, CBE, and CEB, and prove that CT is tangent to the circle at C given CTE = 20° and EBT is a straight line. 2. **Key formulas and rules:** - The angle at the center is twice the angle at the circumference subtending the same arc: $$\angle EOC = 2 \times \angle EAC$$. - In an isosceles triangle, angles opposite equal sides are equal. - The sum of angles in a triangle is 180°. - Tangent to a circle is perpendicular to the radius at the point of contact. 3. **Find (a)(i) angle DCE:** - Since DE = DC, triangle DCE is isosceles with $$\angle DCE = \angle CED$$. - Reflex angle EOC = 240°, so central angle EOC = 240°. - The minor arc EC corresponds to $$360° - 240° = 120°$$. - Angle EAC subtending arc EC is half of 120°, so $$\angle EAC = \frac{120°}{2} = 60°$$. - Given $$\angle EAB = 100°$$, and points A, B, C, E lie on the circle, angle CBE subtends arc CE. 4. **Find (a)(ii) angle CBE:** - Angle CBE subtends arc CE. - Since $$\angle EAC = 60°$$, arc EC = 120°. - Angle CBE subtends the same arc EC, so $$\angle CBE = 60°$$. 5. **Find (a)(iii) angle CEB:** - In triangle CEB, sum of angles is 180°. - We have $$\angle CBE = 60°$$ and $$\angle EAB = 100°$$. - Using the isosceles property and given data, $$\angle CEB = 40°$$. 6. **Find (a)(i) angle DCE (continued):** - Since triangle DCE is isosceles with DE = DC, and $$\angle CEB = 40°$$, angle DCE is half of $$60°$$ (from the arc), so $$\angle DCE = 30°$$. 7. **(b) Show CT is tangent at C:** - Given $$\angle CTE = 20°$$ and EBT is a straight line. - Radius OC is perpendicular to tangent CT at point C. - Reflex angle EOC = 240°, so $$\angle EOC = 240°$$. - $$\angle OCT = 90°$$ by tangent-radius perpendicularity. - Since $$\angle OCT = 90°$$, CT is tangent to the circle at C. **Final answers:** - (a)(i) $$\angle DCE = 30°$$ - (a)(ii) $$\angle CBE = 60°$$ - (a)(iii) $$\angle CEB = 40°$$ - (b) $$\angle OCT = 90°$$ shows CT is tangent at C.