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 $\angle EDF = 40^\circ$ - Angle $\angle FBC = 70^\circ$ - ABC is tangent to the circle at B 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:** Since ABC is tangent at B and FBC is the angle between the tangent and chord BF, then $$\angle FBC = \angle BFE$$ where $\angle BFE$ is the angle in the alternate segment. 5. **Given $\angle FBC = 70^\circ$, so $\angle BFE = 70^\circ$.** 6. **Consider triangle DEF inside the circle:** The sum of angles in triangle DEF is $$\angle EDF + \angle DEF + \angle DFE = 180^\circ$$ 7. **Substitute known angle:** $$40^\circ + \angle DEF + \angle DFE = 180^\circ$$ 8. **Note that $\angle DFE = \angle BFE = 70^\circ$ (since points F, E, B are on the circle and $\angle BFE$ is the same as $\angle DFE$).** 9. **Calculate $\angle DEF$:** $$\angle DEF = 180^\circ - 40^\circ - 70^\circ = 70^\circ$$ 10. **Now, look at angles $\angle DEF$ and $\angle FBC$:** Both are $70^\circ$. 11. **Since $\angle DEF$ and $\angle FBC$ are corresponding angles formed by transversal EF and line ABC, and they are equal, lines ABC and EF are parallel.** 12. **Conclusion:** By the alternate segment theorem and angle properties in triangle DEF, the tangent ABC is parallel to chord EF. **Final answer:** Tangent ABC is parallel to EF because corresponding angles $\angle FBC$ and $\angle DEF$ are equal (both $70^\circ$).