$ \text{Step 1: Given Data } $

The truss is subjected to the following vertical loads:
\begin{array}{c}
W_A = 2\,{\rm{kN}} \quad \text{(at point A)} \\
W_H = 5\,{\rm{kN}} \quad \text{(at point H)} \\
W_G = 5\,{\rm{kN}} \quad \text{(at point G)} \\
W_F = 5\,{\rm{kN}} \quad \text{(at point F)} \\
W_E = 2\,{\rm{kN}} \quad \text{(at point E)}
\end{array}

The truss has the following dimensions:
\begin{array}{c}
AB = 2\,{\rm{m}} \quad \text{(horizontal member)} \\
BC = 2\,{\rm{m}} \quad \text{(horizontal member)} \\
CD = 2\,{\rm{m}} \quad \text{(horizontal member)} \\
DE = 2\,{\rm{m}} \quad \text{(horizontal member)} \\
CG = 3\,{\rm{m}} \quad \text{(vertical member)} \\
AC = 4\,{\rm{m}} \quad \text{(distance from A to C)} \\
BE = 6\,{\rm{m}} \quad \text{(distance from B to E)} \\
AE = 8\,{\rm{m}} \quad \text{(distance from A to E)}
\end{array}

$ \text{Step 2: Determine Support Reactions} $

First, we analyze the entire truss to find the reaction forces:

\begin{array}{c}
\text{Horizontal equilibrium:} \\
\sum F_x = 0 \Rightarrow A_x = 0\,{\rm{kN}} \\
\text{(No horizontal loads are applied, so } A_x \text{ must be zero)}
\end{array}

\begin{array}{c}
\text{Moment equilibrium about point E:} \\
\sum M_E = 0 \Rightarrow -A_y \times AE + W_A \times AE + W_H \times BE + W_G \times CE + W_F \times DE = 0 \\
\text{Where:} \\
CE = AC = 4\,{\rm{m}} \quad \text{(since C is midpoint between A and E)} \\
DE = 2\,{\rm{m}} \quad \text{(given)} \\
\text{Substituting values:} \\
-A_y \times 8 + 2 \times 8 + 5 \times 6 + 5 \times 4 + 5 \times 2 = 0 \\
-8A_y + 16 + 30 + 20 + 10 = 0 \\
\hline
A_y = \frac{16 + 30 + 20 + 10}{8} = \frac{76}{8} = 9.5\,{\rm{kN}}
\end{array}

$ \text{Step 3: Calculate Angles} $

We need to determine the angles for force resolution:

\begin{array}{c}
\text{Angle } \alpha \text{ between members GC and BC:} \\
\tan \alpha = \frac{GC}{BC} = \frac{3\,{\rm{m}}}{2\,{\rm{m}}} = 1.5 \\
\alpha = \tan^{-1}(1.5) \approx 56.3^\circ \\
\text{(This angle is needed to resolve forces at joint B)}
\end{array}

\begin{array}{c}
\text{Angle } \theta \text{ between members GC and AC:} \\
\tan \theta = \frac{GC}{AC} = \frac{3\,{\rm{m}}}{4\,{\rm{m}}} = 0.75 \\
\theta = \tan^{-1}(0.75) \approx 37^\circ \\
\text{(This angle is needed to resolve forces at joint G)}
\end{array}

$ \text{Step 4: Analyze Section ABH} $

We isolate section ABH to find member forces:

\begin{array}{c}
\text{Moment equilibrium about point G to find } F_{BC}: \\
\sum M_G = 0 \Rightarrow W_A \times AC + W_H \times BC + F_{BC} \times GC – A_y \times AC = 0 \\
\text{Where:} \\
AC = 4\,{\rm{m}} \text{ is the horizontal distance} \\
GC = 3\,{\rm{m}} \text{ is the vertical moment arm} \\
\text{Substituting values:} \\
2 \times 4 + 5 \times 2 + F_{BC} \times 3 – 9.5 \times 4 = 0 \\
8 + 10 + 3F_{BC} – 38 = 0 \\
\hline
F_{BC} = \frac{38 – 8 – 10}{3} = \frac{20}{3} \approx 6.67\,{\rm{kN}}\,(\rm{T}) \\
\text{(Positive value indicates tension)}
\end{array}

$ \text{Step 5: Determine Length HB} $

Using geometric similarity of triangles:

\begin{array}{c}
\frac{GC}{HB} = \frac{AC}{BC} \\
\frac{3\,{\rm{m}}}{HB} = \frac{4\,{\rm{m}}}{2\,{\rm{m}}} \\
\hline
HB = \frac{3 \times 2}{4} = 1.5\,{\rm{m}} \\
\text{(This length is needed for moment calculations about point B)}
\end{array}

$ \text{Step 6: Calculate Force in GH} $

\begin{array}{c}
\text{Moment equilibrium about point B:} \\
\sum M_B = 0 \Rightarrow W_A \times AB – A_y \times AB – F_{GH} \times \cos\theta \times HB = 0 \\
\text{Where:} \\
\cos\theta = \cos37^\circ \approx 0.799 \\
HB = 1.5\,{\rm{m}} \text{ (from Step 5)} \\
\text{Substituting values:} \\
2 \times 2 – 9.5 \times 2 – F_{GH} \times 0.799 \times 1.5 = 0 \\
4 – 19 – 1.197F_{GH} = 0 \\
\hline
F_{GH} = \frac{-15}{1.197} \approx -12.5\,{\rm{kN}}\,(\rm{C}) \\
\text{(Negative value indicates compression)}
\end{array}

$ \text{Step 7: Calculate Force in BG} $

\begin{array}{c}
\text{Horizontal force equilibrium at joint B:} \\
\sum F_x = 0 \Rightarrow F_{BC}\cos\alpha + F_{BG} + F_{GH}\cos\theta = 0 \\
\text{Where:} \\
\cos\alpha = \cos56.3^\circ \approx 0.554 \\
\cos\theta = \cos37^\circ \approx 0.799 \\
\text{Substituting values:} \\
6.67 \times 0.554 + F_{BG} – 12.5 \times 0.799 = 0 \\
3.69 + F_{BG} – 9.99 = 0 \\
\hline
F_{BG} = 9.99 – 3.69 = 6.3\,{\rm{kN}}\,(\rm{T}) \\
\text{(Positive value indicates tension)}
\end{array}

$ \text{Step 8: Final Results} $

\begin{array}{c}
\text{Member BC: } 6.67\,{\rm{kN}}\,(\rm{T}) \\
\text{Member GH: } 12.5\,{\rm{kN}}\,(\rm{C}) \\
\text{Member BG: } 6.3\,{\rm{kN}}\,(\rm{T})
\end{array}

$ \text{Step 9: Verification} $

\begin{array}{c}
\text{Moment equilibrium about E: } \sum M_E = 0 \quad \checkmark \\
\text{Horizontal force equilibrium: } \sum F_x = 0 \quad \checkmark \\
\text{Vertical force equilibrium: } \sum F_y = 0 \quad \checkmark \\
\hline
\text{All equilibrium conditions are satisfied}
\end{array}