1. **State the problem:** We are given a geometric figure with points B, F, C, E, and A such that angles $\angle BAE$ and $\angle BCE$ are right angles, and $AE = FE$. Given $AB = 4$ and $BF = 2$, we need to find the length of $CF$. 2. **Analyze the figure and given information:** - $\angle BAE = 90^\circ$ means segment $AE$ is perpendicular to $AB$. - $\angle BCE = 90^\circ$ means segment $CE$ is perpendicular to $BC$. - $AE = FE$ means triangle $AFE$ is isosceles with $AE = FE$. - $AB = 4$ and $BF = 2$ are given lengths. 3. **Set up a coordinate system for clarity:** Place point $B$ at the origin $(0,0)$. Since $AB = 4$ and $\angle BAE = 90^\circ$, place $A$ at $(4,0)$. Since $BF = 2$ and $B$ is at $(0,0)$, place $F$ at $(0,2)$. 4. **Find coordinates of points $E$ and $C$:** Since $AE = FE$ and $A = (4,0)$, $F = (0,2)$, point $E$ lies such that distances $AE = FE$. Let $E = (x,y)$. Distance $AE = \sqrt{(x-4)^2 + (y-0)^2}$. Distance $FE = \sqrt{(x-0)^2 + (y-2)^2}$. Set equal: $$\sqrt{(x-4)^2 + y^2} = \sqrt{x^2 + (y-2)^2}$$ Square both sides: $$(x-4)^2 + y^2 = x^2 + (y-2)^2$$ Expand: $$x^2 - 8x + 16 + y^2 = x^2 + y^2 - 4y + 4$$ Simplify: $$-8x + 16 = -4y + 4$$ Rearranged: $$-8x + 16 + 4y - 4 = 0$$ $$-8x + 4y + 12 = 0$$ Divide by 4: $$-2x + y + 3 = 0$$ So, $$y = 2x - 3$$ 5. **Use right angle $\angle BCE = 90^\circ$:** Points $B = (0,0)$, $C = (c_x, c_y)$, $E = (x,y)$. Since $\angle BCE = 90^\circ$, vectors $CB$ and $CE$ are perpendicular. Vectors: $$\overrightarrow{CB} = (0 - c_x, 0 - c_y) = (-c_x, -c_y)$$ $$\overrightarrow{CE} = (x - c_x, y - c_y)$$ Dot product zero: $$\overrightarrow{CB} \cdot \overrightarrow{CE} = 0$$ $$(-c_x)(x - c_x) + (-c_y)(y - c_y) = 0$$ $$-c_x x + c_x^2 - c_y y + c_y^2 = 0$$ Rearranged: $$c_x^2 + c_y^2 = c_x x + c_y y$$ 6. **Use the fact that $F = (0,2)$ lies on segment $FC$:** We want to find $CF$, so we need coordinates of $C$. 7. **Use the fact that $B, F, C$ are collinear vertically:** Since $B = (0,0)$ and $F = (0,2)$, and $F$ lies between $B$ and $C$, $C$ must be on the vertical line $x=0$. So, $c_x = 0$. 8. **Substitute $c_x = 0$ into the dot product equation:** $$0 + c_y^2 = 0 + c_y y$$ If $c_y \neq 0$, divide both sides by $c_y$: $$c_y = y$$ 9. **So, $C = (0, y)$ and $E = (x, y)$ with $y = 2x - 3$.** 10. **Find length $CF$:** $F = (0,2)$, $C = (0,y)$, so $$CF = |y - 2|$$ 11. **Find $x$ and $y$ using the fact that $E$ lies on segment $FE$ and $AE = FE$:** Recall from step 4: $$y = 2x - 3$$ 12. **Use the fact that $E$ lies on line segment $FE$:** Since $F = (0,2)$ and $E = (x,y)$, and $AE = FE$, we can find $x$ and $y$ by considering the length of $AE$ and $FE$. 13. **Calculate $AE$ and $FE$ in terms of $x$:** $$AE = \sqrt{(x-4)^2 + y^2} = \sqrt{(x-4)^2 + (2x - 3)^2}$$ $$FE = \sqrt{(x-0)^2 + (y-2)^2} = \sqrt{x^2 + (2x - 3 - 2)^2} = \sqrt{x^2 + (2x - 5)^2}$$ 14. **Set $AE = FE$ and square both sides:** $$(x-4)^2 + (2x - 3)^2 = x^2 + (2x - 5)^2$$ Expand: $$(x^2 - 8x + 16) + (4x^2 - 12x + 9) = x^2 + (4x^2 - 20x + 25)$$ Simplify left: $$5x^2 - 20x + 25 = x^2 + 4x^2 - 20x + 25$$ Simplify right: $$5x^2 - 20x + 25 = 5x^2 - 20x + 25$$ Both sides are equal for all $x$, so this confirms the relation but does not determine $x$. 15. **Use the fact that $E$ lies on line segment $BE$:** Since $B = (0,0)$ and $E = (x,y)$, and $\angle BAE = 90^\circ$, point $E$ lies on the line perpendicular to $AB$ at $A$. Since $AB$ is horizontal from $(0,0)$ to $(4,0)$, the perpendicular line at $A$ is vertical at $x=4$. 16. **Therefore, $x = 4$.** 17. **Find $y$:** From step 4: $$y = 2x - 3 = 2(4) - 3 = 8 - 3 = 5$$ 18. **Find $CF$:** $C = (0,y) = (0,5)$, $F = (0,2)$, so $$CF = |5 - 2| = 3$$ **Final answer:** $$\boxed{3}$$