In riangle ACE, BF ext is parallel to CE, and B lies on AC, F lies on AE. If BC = rac38 AC, what is the ratio racABAC?

Step 1: Use the parallel lines $BF \parallel CE$ to establish ratios.
Given $BF \parallel CE$ in $\triangle ACE$.
By the Basic Proportionality Theorem (or similar triangles $\triangle ABF \sim \triangle ACE$), we have:

$$\frac{AB}{AC} = \frac{AF}{AE} = \frac{BF}{CE}$$

We are given $BC = \frac{3}{8} AC$.
Since $AB + BC = AC$, we can write $AB = AC - BC = AC - \frac{3}{8} AC = \frac{5}{8} AC$.
Therefore, the ratio $\frac{AB}{AC} = \frac{5}{8}$.
This implies $\frac{