1. **Stating the problem:** We have triangle HFG with points D on HF and E on HG such that DE is parallel to FG. Given: |DH| = 5 cm, |HE| = 12 cm, |HG| = 30 cm, and the distance from D to F is $x$ cm. We need to find the value of $x$. 2. **Formula and rule:** When a segment DE is drawn parallel to one side FG of triangle HFG, it creates two similar triangles: triangle HDE and triangle HFG. By the properties of similar triangles, corresponding sides are proportional: $$\frac{|DH|}{|HF|} = \frac{|HE|}{|HG|} = \frac{|DE|}{|FG|}$$ 3. **Using the given lengths:** We know |DH| = 5 cm, |HE| = 12 cm, and |HG| = 30 cm. Since D lies on HF, and we want to find |DF| = $x$, then: $$|HF| = |DH| + |DF| = 5 + x$$ 4. **Setting up the proportion:** Using the similarity ratio: $$\frac{|DH|}{|HF|} = \frac{|HE|}{|HG|}$$ Substitute known values: $$\frac{5}{5 + x} = \frac{12}{30}$$ 5. **Solving for $x$:** Cross-multiply: $$5 \times 30 = 12 \times (5 + x)$$ $$150 = 60 + 12x$$ Subtract 60 from both sides: $$150 - 60 = 12x$$ $$90 = 12x$$ Divide both sides by 12: $$\frac{90}{\cancel{12}} = \frac{12x}{\cancel{12}}$$ $$7.5 = x$$ 6. **Final answer:** The distance from D to F is $\boxed{7.5}$ cm.