In this problem, we're given a triangle $\Delta ABC$, with points $D$ and $E$ lying on sides $AB$ and $AC$ respectively, such that $DE \parallel BC$. We are also given:

• $AD = \frac{1}{3} BD$
• $AE = 4.5 \, \text{cm}$

We are asked to find $AC$.

Step-by-Step Solution:

1. Use the Basic Proportionality Theorem (Thales' Theorem): Since $DE \parallel BC$, by the Basic Proportionality Theorem, the segments along sides $AB$ and $AC$ are proportional. This means: $\frac{AD}{BD} = \frac{AE}{EC}$ We are given $AD = \frac{1}{3} BD$, which implies: $\frac{AD}{BD} = \frac{1}{3}$ Hence: $\frac{AE}{EC} = \frac{1}{3}$

2. Calculate the Length of $EC$: We are given $AE = 4.5 \, \text{cm}$. Using the ratio $\frac{AE}{EC} = \frac{1}{3}$, we can express $EC$ as: $AE = \frac{1}{3} EC$ Therefore: $EC = 3 \times AE = 3 \times 4.5 = 13.5 \, \text{cm}$

3. Find the Total Length of $AC$: The total length of $AC$ is the sum of $AE$ and $EC$: $AC = AE + EC = 4.5 + 13.5 = 18 \, \text{cm}$

Thus, the length of $AC$ is $18 \, \text{cm}$.

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Would you like further clarification on any part of the solution?

Here are five related questions for deeper understanding:

1. How does the Basic Proportionality Theorem apply in triangles with parallel lines?
2. If $AD = \frac{1}{4} BD$, how would the length of $AC$ change?
3. What if $AE$ was unknown, and only $AC$ was given? How would you solve it?
4. Can the Basic Proportionality Theorem be used in non-triangular shapes?
5. How would you prove that $DE \parallel BC$?

Tip: When dealing with proportionality in geometry, always check for parallel lines, as they often imply a proportional relationship between segments.