$ \text{Step 1: Problem Statement and Given Data} $

The truss is subjected to the following loads:
\begin{array}{c}
F_1 = 4\,{\rm{kN}} \quad \text{(at point E)} \\
F_2 = 8\,{\rm{kN}} \quad \text{(at point F)}
\end{array}

Key dimensions:
\begin{array}{c}
ED = a = 1.5\,{\rm{m}} \\
CD = b = 2\,{\rm{m}} \\
BC = c = 2\,{\rm{m}}
\end{array}

$ \text{Step 2: Free Body Diagram} $

The free body diagram shows:

\begin{array}{c}
\text{Member AF: Force } F_{AF} \\
\text{Member BF: Force } F_{BF} \\
\text{Member BC: Force } F_{BC}
\end{array}

$ \text{Step 3: Force in BC (Moment Equilibrium about F)} $

\begin{array}{c}
\sum M_F = 0 \\
-F_1 \times b – F_{BC} \times a = 0 \\
\text{Where:} \\
b = 2\,{\rm{m}} \text{ (moment arm for } F_1) \\
a = 1.5\,{\rm{m}} \text{ (moment arm for BC)} \\
\text{Substituting values:} \\
-4 \times 2 – F_{BC} \times 1.5 = 0 \\
-8 – 1.5F_{BC} = 0 \\
\hline
F_{BC} = \frac{-8}{1.5} \approx -5.33\,{\rm{kN}}\,(\rm{C}) \\
\text{(Negative value indicates compression)}
\end{array}

$ \text{Step 4: Length of BF (Pythagorean Theorem)} $

\begin{array}{c}
BF = \sqrt{a^2 + c^2} \\
= \sqrt{(1.5)^2 + (2)^2} \\
= \sqrt{2.25 + 4} \\
\hline
BF = 2.5\,{\rm{m}}
\end{array}

$ \text{Step 5: Force in BF (Horizontal Equilibrium)} $

\begin{array}{c}
\sum F_x = 0 \\
F_{BF}\cos\theta + F_1 + F_2 = 0 \\
\text{Where:} \\
\cos\theta = \frac{a}{BF} = \frac{1.5}{2.5} = 0.6 \\
\text{Substituting values:} \\
F_{BF} \times 0.6 + 4 + 8 = 0 \\
0.6F_{BF} = -12 \\
\hline
F_{BF} = \frac{-12}{0.6} = -20\,{\rm{kN}}\,(\rm{C}) \\
\text{(Negative value indicates compression)}
\end{array}

$ \text{Step 6: Force in AF (Vertical Equilibrium)} $

\begin{array}{c}
\sum F_y = 0 \\
-F_{AF} – F_{BC} – F_{BF}\sin\theta = 0 \\
\text{Where:} \\
\sin\theta = \frac{c}{BF} = \frac{2}{2.5} = 0.8 \\
\text{Substituting values:} \\
-F_{AF} – (-5.33) – (-20)(0.8) = 0 \\
-F_{AF} + 5.33 + 16 = 0 \\
\hline
F_{AF} = 21.33\,{\rm{kN}}\,(\rm{T}) \\
\text{(Positive value indicates tension)}
\end{array}

$ \text{Step 7: Verification} $

\begin{array}{c}
\text{Moment Check: } \sum M_F = 0 \quad \checkmark \\
\text{Force Balance: } \sum F_x = \sum F_y = 0 \quad \checkmark \\
\hline
\text{All equilibrium conditions are satisfied}
\end{array}

$ \text{Step 8: Final Results} $

\begin{array}{|c|c|c|}
\hline
\text{Member} & \text{Force (kN)} & \text{Nature} \\
\hline
BC & 5.33 & Compression \\
BF & 20 & Compression \\
AF & 21.33 & Tension \\
\hline
\end{array}