1. **Problem statement:** We have a circle with center O and points A, B, C on the circumference. CD is a tangent at C. Given $\angle BCD = 40^\circ$ and $\angle OAB = 3 \times \angle OAC$, we need to find $\angle ACD$ and state the circle theorems used. 2. **Circle theorems used:** - The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment. - The angle at the center is twice the angle at the circumference subtending the same arc. 3. **Step 1: Use the tangent-chord theorem** Since CD is tangent at C and BCD is given, by the tangent-chord theorem: $$\angle BCD = \angle BAC = 40^\circ$$ 4. **Step 2: Express angles at A** Let $\angle OAC = x$. Then $\angle OAB = 3x$ (given). 5. **Step 3: Use the fact that $\angle BAC = \angle OAB + \angle OAC$** Since A, B, C are points on the circle, $\angle BAC = \angle OAB + \angle OAC = 3x + x = 4x$. 6. **Step 4: Equate to find x** We know $\angle BAC = 40^\circ$, so: $$4x = 40^\circ$$ $$x = 10^\circ$$ 7. **Step 5: Find $\angle OAB = 3x = 30^\circ$ and $\angle OAC = 10^\circ$** 8. **Step 6: Use the angle at center theorem** The angle at the center $\angle AOB$ is twice the angle at the circumference $\angle ACB$ subtending the same arc AB: $$\angle AOB = 2 \times \angle ACB$$ 9. **Step 7: Find $\angle AOB$** Since O is center, $\triangle OAB$ is isosceles with OA = OB (radii). Angles at A and B are $\angle OAB = 30^\circ$ and $\angle OBA = 30^\circ$ (by symmetry), so: $$\angle AOB = 180^\circ - 30^\circ - 30^\circ = 120^\circ$$ 10. **Step 8: Find $\angle ACB$** $$\angle ACB = \frac{\angle AOB}{2} = \frac{120^\circ}{2} = 60^\circ$$ 11. **Step 9: Find $\angle ACD$** In $\triangle BCD$, $\angle BCD = 40^\circ$ (given), and $\angle ACB = 60^\circ$ (found). Since points A, C, D are on the circle and tangent, $\angle ACD = \angle ACB = 60^\circ$ (alternate segment theorem). **Final answer:** $$\boxed{\angle ACD = 60^\circ}$$