1. **State the problem:** We have a triangle L C X with points B on L C and D on C X. Segment B D is 3 m, B C is 7 m, D X is 1.4 m, and X Y is 18 m. We need to find the values of $x$ and $y$. 2. **Identify similar triangles and use ratios:** Since B and D lie on sides L C and C X respectively, and B D is drawn, triangles L B D and L X Y are similar by the AA criterion (corresponding angles equal). 3. **Set up ratios from similar triangles:** The sides correspond as follows: - $\frac{B C}{C X} = \frac{x}{1.4}$ - $\frac{B D}{X Y} = \frac{3}{18}$ 4. **Calculate $x$ using the ratio:** $$\frac{x}{1.4} = \frac{3}{18}$$ Multiply both sides by 1.4: $$x = 1.4 \times \frac{3}{18}$$ Simplify: $$x = 1.4 \times \frac{1}{6} = \frac{1.4}{6}$$ Cancel common factors: $$x = \frac{\cancel{1.4}}{6} = 0.2333...$$ So, $$x \approx 0.233$$ meters. 5. **Calculate $y$ using the total length:** Since $y$ is the segment from B to Y along the line, and $X Y = 18$ m, and $B D = 3$ m, the total length $y$ is: $$y = B D + D X + X Y$$ Given $D X = 1.4$ m and $B D = 3$ m, but $y$ is the segment from B to Y, so: $$y = 3 + 18 = 21$$ meters. 6. **Final answers:** $$x \approx 0.233 \text{ m}$$ $$y = 21 \text{ m}$$