$ \text{Step 1: 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 is constructed for section $a-a$, showing:

\begin{array}{c}
\text{Member EF: Force } F_{EF} \\
\text{Member CF: Force } F_{CF} \\
\text{Member BC: Force } F_{BC}
\end{array}

$ \text{Step 3: Force in EF (Moment Equilibrium about C)} $

\begin{array}{c}
\sum M_C = 0 \\
F_{EF} \times a – F_1 \times b = 0 \\
\text{Where:} \\
a = 1.5\,{\rm{m}} \text{ (moment arm for EF)} \\
b = 2\,{\rm{m}} \text{ (moment arm for } F_1) \\
\text{Substituting values:} \\
F_{EF} \times 1.5 – 4 \times 2 = 0 \\
1.5F_{EF} = 8 \\
\hline
F_{EF} = \frac{8}{1.5} \approx 5.33\,{\rm{kN}}\,(\rm{T}) \\
\text{(Positive value indicates tension)}
\end{array}

$ \text{Step 4: Force in BC (Vertical Equilibrium)} $

\begin{array}{c}
\sum F_y = 0 \\
-F_{BC} – F_{EF} = 0 \\
\text{Using } F_{EF} = 5.33\,{\rm{kN}} \text{ from Step 3:} \\
-F_{BC} – 5.33 = 0 \\
\hline
F_{BC} = -5.33\,{\rm{kN}}\,(\rm{C}) \\
\text{(Negative value indicates compression)}
\end{array}

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

\begin{array}{c}
\sum F_x = 0 \\
F_1 + F_2 – F_{CF} = 0 \\
\text{Substituting given forces:} \\
4 + 8 – F_{CF} = 0 \\
\hline
F_{CF} = 12\,{\rm{kN}}\,(\rm{T}) \\
\text{(Positive value indicates tension)}
\end{array}

$ \text{Step 6: Verification} $

\begin{array}{c}
\text{Moment Check: } \sum M_C = 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 7: Final Results} $

\begin{array}{|c|c|c|}
\hline
\text{Member} & \text{Force (kN)} & \text{Nature} \\
\hline
EF & 5.33 & Tension \\
BC & 5.33 & Compression \\
CF & 12 & Tension \\
\hline
\end{array}