1. **Problem statement:** Given triangle ABCD with point E on side AD such that BE is parallel to DC, and lengths AB = x, BC = 6, AC = 7, CD = 16, find the value of $x$. 2. **Key concept:** When a line (BE) is drawn parallel to one side (DC) of a triangle, it divides the other two sides proportionally. This is the Basic Proportionality Theorem (Thales' theorem). 3. **Apply the theorem:** Since BE is parallel to DC, the segments on sides AB and AC are proportional: $$\frac{AB}{BC} = \frac{AE}{ED}$$ 4. **Identify segments:** We know: - $AB = x$ - $BC = 6$ - $AC = 7$ - $CD = 16$ Since E lies on AD, and AD = AE + ED, but AD is not given directly. However, since BE is parallel to DC, the ratio $\frac{AB}{BC} = \frac{AE}{ED}$. 5. **Express AE and ED:** Let $AE = y$, then $ED = AD - y$. But AD is unknown. However, since BE divides AC and AD proportionally, and given the lengths, the ratio $\frac{AB}{BC} = \frac{AE}{ED}$ can be used with the known lengths. 6. **Use the proportionality on AC and AD:** Since BE divides AC and AD proportionally, the ratio of segments on AC and AD are equal: $$\frac{AB}{BC} = \frac{AE}{ED} = \frac{AC}{CD}$$ 7. **Calculate ratio $\frac{AC}{CD}$:** $$\frac{AC}{CD} = \frac{7}{16}$$ 8. **Set up equation:** $$\frac{x}{6} = \frac{7}{16}$$ 9. **Solve for $x$:** $$x = 6 \times \frac{7}{16}$$ $$x = \frac{42}{16}$$ 10. **Simplify fraction:** $$x = \frac{\cancel{42}^{21}}{\cancel{16}^{8}} = \frac{21}{8} = 2.625$$ **Final answer:** $$x = \frac{21}{8} = 2.625$$