1. **Stating the problem:** We are given a triangle ABC with points D, E, F, G, H, and I extending from it. We know angles AFC = $x^\circ$, ACB = $y^\circ$, BCF = $40^\circ$, and AFE = $50^\circ$. We need to find the values of $x$ and $y$. 2. **Analyzing the given information:** The problem mentions that segments BF, FD, and CI have equal lengths (indicated by single tick marks). Also, CF is extended to G, AB extended beyond B to H, and DE beyond E to I. The base BE is a straight horizontal line. 3. **Using angle relationships:** Since BF = FD, triangle BFD is isosceles with base BD. Therefore, angles at B and D in triangle BFD are equal. 4. **Finding $x$ (angle AFC):** Angle AFC is formed at point F by points A and C. Given angle AFE = $50^\circ$ and BCF = $40^\circ$, and considering the isosceles triangles and linear pairs, we can deduce: - Since BF = FD, angles BFD and BDF are equal. - Angle BCF = $40^\circ$ and angle AFE = $50^\circ$ are adjacent to angle AFC. By the exterior angle theorem and linear pairs, angle AFC $x = 180^\circ - (50^\circ + 40^\circ) = 90^\circ$. 5. **Finding $y$ (angle ACB):** Angle ACB is at point C in triangle ABC. Since CF is extended to G and considering the equal lengths and angles, angle ACB $y$ is equal to angle BCF plus angle AFC: $$y = 40^\circ + 90^\circ = 130^\circ$$ 6. **Final answers:** $$x = 90^\circ$$ $$y = 130^\circ$$