1. **Problem Statement:** Given triangle ABC with vertices B(5,0), C(1,4), and A unknown. D is the midpoint of AC and lies on the y-axis. AE is the perpendicular bisector of BC and intersects BD at F. 2. **Find the coordinates of E, the midpoint of BC:** $$E = \left(\frac{5+1}{2}, \frac{0+4}{2}\right) = (3, 2)$$ 3. **Find the slope of BC:** $$m_{BC} = \frac{4-0}{1-5} = \frac{4}{-4} = -1$$ 4. **Slope of AE (perpendicular bisector of BC):** Since AE is perpendicular to BC, $$m_{AE} = -\frac{1}{m_{BC}} = -\frac{1}{-1} = 1$$ 5. **Find coordinates of A:** D is midpoint of AC and lies on y-axis, so its x-coordinate is 0. Let A = (x, y), C = (1,4), D = midpoint of AC = \left(\frac{x+1}{2}, \frac{y+4}{2}\right) Since D lies on y-axis, its x-coordinate is 0: $$\frac{x+1}{2} = 0 \implies x = -1$$ 6. **Find y-coordinate of D:** $$D_y = \frac{y+4}{2}$$ 7. **Find equation of AE passing through E(3,2) with slope 1:** $$y - 2 = 1(x - 3) \implies y = x - 1$$ 8. **Find equation of BD:** B = (5,0), D = \left(0, \frac{y+4}{2}\right) Slope of BD: $$m_{BD} = \frac{\frac{y+4}{2} - 0}{0 - 5} = \frac{y+4}{2 \times (-5)} = -\frac{y+4}{10}$$ Equation of BD: $$y - 0 = m_{BD}(x - 5) \implies y = -\frac{y+4}{10}(x - 5)$$ 9. **Find F, intersection of AE and BD:** From AE: $$y = x - 1$$ From BD: $$y = -\frac{y+4}{10}(x - 5)$$ Set equal: $$x - 1 = -\frac{y+4}{10}(x - 5)$$ 10. **Find y-coordinate of A by using D lies on y-axis:** Since D is midpoint of AC and lies on y-axis, D = (0, d_y) with $$d_y = \frac{y+4}{2}$$ 11. **Use D lies on y-axis and AE passes through A:** A lies on AE, slope 1, passing through E(3,2), so A satisfies: $$y - 2 = 1(x - 3) \implies y = x - 1$$ Since A = (-1, y), substitute x = -1: $$y = -1 - 1 = -2$$ 12. **Coordinates of A:** $$A = (-1, -2)$$ 13. **Coordinates of D:** $$D = \left(0, \frac{-2 + 4}{2}\right) = (0, 1)$$ 14. **Equation of BD:** Slope: $$m_{BD} = \frac{1 - 0}{0 - 5} = \frac{1}{-5} = -\frac{1}{5}$$ Equation: $$y - 0 = -\frac{1}{5}(x - 5) \implies y = -\frac{1}{5}x + 1$$ 15. **Find coordinates of F (intersection of AE and BD):** Set AE and BD equal: $$x - 1 = -\frac{1}{5}x + 1$$ Multiply both sides by 5: $$5x - 5 = -x + 5$$ $$5x + x = 5 + 5$$ $$6x = 10 \implies x = \frac{10}{6} = \frac{5}{3}$$ Find y: $$y = x - 1 = \frac{5}{3} - 1 = \frac{2}{3}$$ Coordinates of F: $$F = \left(\frac{5}{3}, \frac{2}{3}\right)$$ 16. **Find areas of triangles FAB and FBE:** Vertices: $$F = \left(\frac{5}{3}, \frac{2}{3}\right), A = (-1, -2), B = (5, 0), E = (3, 2)$$ Area formula for triangle with points $(x_1,y_1), (x_2,y_2), (x_3,y_3)$: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ 17. **Area of $\triangle FAB$:** $$= \frac{1}{2} | \frac{5}{3}(-2 - 0) + (-1)(0 - \frac{2}{3}) + 5(\frac{2}{3} + 2) |$$ Calculate inside: $$= \frac{1}{2} | \frac{5}{3}(-2) + (-1)(-\frac{2}{3}) + 5(\frac{8}{3}) |$$ $$= \frac{1}{2} | -\frac{10}{3} + \frac{2}{3} + \frac{40}{3} | = \frac{1}{2} | \frac{32}{3} | = \frac{16}{3}$$ 18. **Area of $\triangle FBE$:** $$= \frac{1}{2} | \frac{5}{3}(0 - 2) + 5(2 - \frac{2}{3}) + 3(\frac{2}{3} - 0) |$$ Calculate inside: $$= \frac{1}{2} | \frac{5}{3}(-2) + 5(\frac{4}{3}) + 3(\frac{2}{3}) |$$ $$= \frac{1}{2} | -\frac{10}{3} + \frac{20}{3} + 2 | = \frac{1}{2} | \frac{10}{3} + 2 | = \frac{1}{2} \times \frac{16}{3} = \frac{8}{3}$$ **Final answers:** (a) Equation of AE: $$y = x - 1$$ (b)(i) Coordinates of A: $$(-1, -2)$$ (b)(ii) Equation of BD: $$y = -\frac{1}{5}x + 1$$ (c)(i) Coordinates of F: $$\left(\frac{5}{3}, \frac{2}{3}\right)$$ (c)(ii) Area of $\triangle FAB = \frac{16}{3}$ Area of $\triangle FBE = \frac{8}{3}$