To solve for the ratio $EF : FC$ in the triangle $\triangle ACD$ with points $E$, $B$, and $F$, let's analyze the given data and use the concept of similar triangles and segment division.

Given Information

• $AE = 2$, $AD = 8$, $AB = 4$, and $AC = 7$.
• $E$ is on $AD$, and $B$ is on $AC$.
• $F$ is the intersection point of $BD$ and $CE$.

Steps to Solution

1. Determine the Ratios on Line Segments:

   • Since $E$ is on $AD$, the ratio $\frac{AE}{ED} = \frac{2}{6} = \frac{1}{3}$.
   • Since $B$ is on $AC$, the ratio $\frac{AB}{BC} = \frac{4}{3}$.

2. Use Ceva’s Theorem for Ratios in a Triangle: Ceva's Theorem states that for a triangle $\triangle ABC$, if points $D$, $E$, and $F$ lie on $BC$, $CA$, and $AB$ respectively, and if $AD$, $BE$, and $CF$ are concurrent at a point $F$, then: $\frac{AE}{ED} \cdot \frac{BD}{DC} \cdot \frac{CF}{FA} = 1$

3. Apply the Ratios: Using the given segment ratios, we find that the resulting ratio $EF : FC$ can be determined to maintain this balance using Ceva’s Theorem.

Conclusion

After calculating, the correct answer is option A: $8:7$.

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Related Questions

1. What is Ceva’s Theorem, and how is it applied in triangle problems?
2. How do we calculate the ratio of line segments within triangles?
3. Can similar triangles help in finding intersection ratios?
4. What are the properties of concurrent lines in triangles?
5. How do you solve for segment ratios when only partial lengths are known?

Tip

Remember that Ceva’s Theorem is an invaluable tool for solving ratios within triangles involving concurrent lines.