1. **Problem statement:** We need to solve for the value of $x$ and find the lengths of segments $AD$ and $DB$ in the given triangle with sides labeled as $BD = 2x$, $DA = x + 2$, $CE = 32$, and $EA = 24$. 2. **Understanding the problem:** Since $BD$ and $DA$ are parts of segment $BA$, and $CE$ and $EA$ are parts of segment $CA$, we can use the segment addition postulate and properties of similar triangles if applicable. 3. **Assuming triangles are similar:** If triangles $CBE$ and $DEA$ are similar (common in such problems), then the ratios of corresponding sides are equal: $$\frac{BD}{DA} = \frac{CE}{EA}$$ 4. **Substitute the given lengths:** $$\frac{2x}{x+2} = \frac{32}{24}$$ 5. **Simplify the right side:** $$\frac{32}{24} = \frac{4}{3}$$ 6. **Set up the equation:** $$\frac{2x}{x+2} = \frac{4}{3}$$ 7. **Cross multiply:** $$3 \times 2x = 4 \times (x+2)$$ $$6x = 4x + 8$$ 8. **Solve for $x$:** $$6x - 4x = 8$$ $$2x = 8$$ $$x = 4$$ 9. **Find lengths $AD$ and $DB$:** $$AD = x + 2 = 4 + 2 = 6$$ $$DB = 2x = 2 \times 4 = 8$$ **Final answer:** $$x = 4, \quad AD = 6, \quad DB = 8$$