1. **Problem Statement:** Given triangle ABC with segment DE parallel to BC and segment EF parallel to AB, and given lengths BC = 24, AC = 18, AB = 18, DE = 6, EF = 12, determine which statement about segments BD or BF is true using the theorem that a line parallel to one side of a triangle divides the other two sides proportionately. 2. **Theorem Used:** If a line is parallel to one side of a triangle, it divides the other two sides proportionally. For example, if DE \parallel BC, then \frac{AD}{DB} = \frac{AE}{EC}. 3. **Apply the theorem to DE \parallel BC:** Since DE \parallel BC, the segments on sides AB and AC are divided proportionally: $$\frac{AD}{DB} = \frac{AE}{EC} = \frac{DE}{BC}$$ Given: - $DE = 6$ - $BC = 24$ So, $$\frac{DE}{BC} = \frac{6}{24} = \frac{1}{4}$$ 4. **Find BD:** Since $AB = 18$, let $AD = x$ and $DB = 18 - x$. From the proportion: $$\frac{AD}{DB} = \frac{1}{4}$$ Cross-multiplied: $$4AD = DB$$ Substitute $DB = 18 - AD$: $$4AD = 18 - AD$$ $$4AD + AD = 18$$ $$5AD = 18$$ $$AD = \frac{18}{5} = 3.6$$ Then, $$DB = 18 - 3.6 = 14.4$$ 5. **Apply the theorem to EF \parallel AB:** Since EF \parallel AB, the segments on sides DE and DC are divided proportionally. However, since EF is parallel to AB, and EF = 12, AB = 18, the ratio is: $$\frac{EF}{AB} = \frac{12}{18} = \frac{2}{3}$$ 6. **Find BF:** Since $BC = 24$, and EF \parallel AB divides BC proportionally, let $BF = y$ and $FC = 24 - y$. From the theorem: $$\frac{BF}{FC} = \frac{EF}{AB} = \frac{2}{3}$$ Cross-multiplied: $$3BF = 2FC$$ Substitute $FC = 24 - BF$: $$3BF = 2(24 - BF)$$ $$3BF = 48 - 2BF$$ $$3BF + 2BF = 48$$ $$5BF = 48$$ $$BF = \frac{48}{5} = 9.6$$ 7. **Compare with given options:** - Segment BD = 12 (No, calculated 14.4) - Segment BD = 4 (No) - Segment BF = 16 (No, calculated 9.6) - Segment BF = 9 (Closest to 9.6, so approximately true) **Final answer:** Segment BF = 9 is the statement that can be proved approximately true using the theorem.