1. **State the problem:** We need to find the values of $x$ and $y$ in the given geometric figure involving triangle $ACX$ with points $B$, $D$, and $F$ on the sides and segments as described. 2. **Identify given information and relationships:** - $AB \parallel XF$ - $XF = 21$ m - $CB = 10$ m - $BD = 8$ m - $DX = 18$ m - $CD = x$ - $DF = y$ 3. **Use properties of parallel lines and similar triangles:** Since $AB \parallel XF$, triangles $ABC$ and $XFC$ are similar by the AA criterion. 4. **Set up ratios from similar triangles:** $$\frac{AB}{XF} = \frac{CB}{CF}$$ We know $AB$ is parallel to $XF$ and $XF=21$ m. 5. **Find $CF$:** Segment $CF$ is composed of $CD + DF = x + y$. 6. **Use the segment $BD$ and $DX$ to find $x$:** Since $B$ lies on $AC$ and $D$ lies on $CX$, and $BD$ extends to meet $CF$, triangles $CBD$ and $XDF$ are similar. 7. **Set up similarity ratio for triangles $CBD$ and $XDF$:** $$\frac{CB}{XD} = \frac{BD}{DF} = \frac{CD}{XF}$$ Given $CB=10$, $BD=8$, $DX=18$, $XF=21$, and $DF=y$, $CD=x$. 8. **Calculate $y$ using ratio $\frac{BD}{DF} = \frac{CB}{XD}$:** $$\frac{8}{y} = \frac{10}{18} \Rightarrow 8 \times 18 = 10 \times y \Rightarrow 144 = 10y \Rightarrow y = \frac{144}{10} = 14.4$$ 9. **Calculate $x$ using ratio $\frac{CD}{XF} = \frac{CB}{XD}$:** $$\frac{x}{21} = \frac{10}{18} \Rightarrow x = 21 \times \frac{10}{18} = \frac{210}{18} = 11.67$$ 10. **Final answers:** $$x = 11.67 \text{ m}, \quad y = 14.4 \text{ m}$$