To solve for $x$ in this triangle problem, we can use the concept of similar triangles. The given triangle is split by a line segment that is parallel to the base, creating two smaller triangles that are similar to the larger triangle.

Step-by-step solution:

1. Identify the triangles:

   • The two triangles formed are $\triangle ABD$ and $\triangle ABC$, where $\triangle ABD \sim \triangle ABC$.

2. Write the similarity ratio: Since $DE$ is parallel to $BC$, the ratios of corresponding sides are equal: $\frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE}.$

3. Focus on segment ratios:

   • From the given figure:
     • $BC = 12$ (entire base of $\triangle ABC$),
     • $BE = x$,
     • $EC = 9$,
     • $BD = x$,
     • $DE = 3$.

4. Use the segment proportionality theorem: The ratios of the segments on $BC$ (the parallel line divides the base proportionally): $\frac{BD}{BE} = \frac{DE}{BC}.$

5. Substitute known values:

   • $BD = x$,
   • $BE = x + 9$,
   • $DE = 3$,
   • $BC = 12$.

   Substitute into the proportion: $\frac{x}{x + 9} = \frac{3}{12}.$

6. Simplify the equation:

   • Simplify $\frac{3}{12}$ to $\frac{1}{4}$: $\frac{x}{x + 9} = \frac{1}{4}.$

   • Cross-multiply: $4x = x + 9.$

   • Solve for $x$: $4x - x = 9,$ $3x = 9,$ $x = 3.$

Final Answer:

$x = 3.$

Let me know if you'd like a detailed explanation or have questions. Here are 5 related questions to deepen your understanding:

1. What are the conditions for triangles to be similar?
2. How does the parallel line theorem apply in geometric proofs?
3. Can you solve for $x$ using a different ratio of similar triangles?
4. How would this solution change if the line segment $DE$ was not parallel to $BC$?
5. What are other real-world applications of triangle similarity?

Tip: Always look