1. **State the problem:** We need to prove that the tangent line ABC at point B is parallel to the chord EF of the circle. 2. **Given:** - Angle EDF = 40° - Angle FBC = 70° - ABC is tangent to the circle at B - Points B, D, E, F lie on the circle 3. **Recall the tangent-chord angle theorem:** The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment of the circle. 4. **Apply the theorem:** - The angle between tangent ABC and chord BF is angle FBC = 70° (given). - By the tangent-chord theorem, angle FBC = angle BDE (angle in the alternate segment). 5. **Find angle BDE:** - Triangle EDF has angles EDF = 40° (given), and we want to find angle DEF. - Since points D, E, F lie on the circle, angle DEF is the angle subtended by chord DF at point E. 6. **Use the fact that angles subtended by the same chord are equal:** - Angle EDF = 40° (given) - Angle DEF = angle BDE (since B, D, E lie on the circle and subtend the same chord) 7. **Calculate angle DEF:** - Sum of angles in triangle DEF is 180° - Let angle DEF = x - Then, $x + 40° + \text{angle DFE} = 180°$ 8. **Use the fact that angle FBC = 70° equals angle BDE:** - So angle BDE = 70° 9. **Since angle FBC = angle BDE, the tangent-chord theorem confirms that ABC is parallel to EF because corresponding angles are equal.** 10. **Conclusion:** - Tangent ABC is parallel to chord EF because the alternate segment angles are equal (70°), satisfying the condition for parallel lines. **Final answer:** Tangent ABC is parallel to EF by the tangent-chord angle theorem and equality of alternate segment angles.