1. **Problem statement:** We have points A, B, C, D, and E on a circle, and FG is a tangent to the circle at point C. We know these angles: - Angle BAD = 110° - Angle ADB = 20° - Angle BEC = 45° We need to find angle BCD and explain why. 2. **Important rule:** In a circle, the opposite angles of a cyclic quadrilateral add up to 180°. Also, the angle between a tangent and a chord equals the angle in the alternate segment. 3. **Step 1: Find angle ABD.** Since triangle ABD has angles BAD = 110° and ADB = 20°, the third angle ABD is: $$\text{Angle ABD} = 180^\circ - 110^\circ - 20^\circ = 50^\circ$$ 4. **Step 2: Use the cyclic quadrilateral property.** Points A, B, C, D lie on the circle, so quadrilateral ABCD is cyclic. Opposite angles add to 180°, so: $$\text{Angle BAD} + \text{Angle BCD} = 180^\circ$$ We know angle BAD = 110°, so: $$\text{Angle BCD} = 180^\circ - 110^\circ = 70^\circ$$ 5. **Step 3: Geometrical reason using tangent.** The tangent FG at C makes an angle with chord BC equal to the angle in the alternate segment, which is angle BCD. Since angle BEC = 45° is given, and E lies on the circle, this confirms the relationship between tangent and alternate segment angles. **Final answer:** Angle BCD = 70° because the opposite angles in cyclic quadrilateral ABCD add up to 180°, and the tangent FG at C creates an angle equal to angle BCD by the alternate segment theorem.