$ \text{Step 1:  Given Data} $

A truss structure is subjected to five vertical loads of $1500\,{\rm{lb}} $ each at points B through F. Key dimensions:
\begin{array}{c}
{EF} = 3\,{\rm{ft}} \\
{GE} = 4\,{\rm{ft}} \\
{DE} = 3\,{\rm{ft}} \\
{HD} = 4\,{\rm{ft}}
\end{array}

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

Analyze section EFG to determine forces in:
\begin{array}{c}
\text{Member CD: } F_{CD} \\
\text{Member HI: } F_{HI} \\
\text{Member CH: } F_{CH}
\end{array}

$ \text{Step 3: Geometric Analysis} $

Calculate angle $\theta$ for member CH:
\begin{array}{c}
\tan \theta = \frac{GE}{EF} = \frac{4}{3} \\
\theta = \tan^{-1}\left(\frac{4}{3}\right) \\
\hline
\theta \approx 53.1^\circ
\end{array}

$ \text{Step 4: Force in Member CD} $

Moment equilibrium about H:
\begin{array}{c}
\sum M_H = 0 \\
-F_{CD}(4\,{\rm{ft}}) – 1500\,{\rm{lb}}(3\,{\rm{ft}}) – 1500\,{\rm{lb}}(6\,{\rm{ft}}) = 0 \\
-4F_{CD} = 13500\,{\rm{lb}\cdot{\rm{ft}}} \\
\hline
F_{CD} = -3375\,{\rm{lb}}\,(\rm{C})
\end{array}

$ \text{Step 5: Force in Member CH} $

Vertical equilibrium at C:
\begin{array}{c}
\sum F_y = 0 \\
-F_{CH}\sin(53.1^\circ) – 4500\,{\rm{lb}} = 0 \\
F_{CH} = \frac{-4500\,{\rm{lb}}}{0.8} \\
\hline
F_{CH} = -5625\,{\rm{lb}}\,(\rm{C})
\end{array}

$ \text{Step 6: Force in Member HI} $

Horizontal equilibrium at H:
\begin{array}{c}
\sum F_x = 0 \\
-F_{HI} – (-3375\,{\rm{lb}}) – (-5625\,{\rm{lb}})\cos(53.1^\circ) = 0 \\
F_{HI} = 3375\,{\rm{lb}} + 3375\,{\rm{lb}} \\
\hline
F_{HI} = 6750\,{\rm{lb}}\,(\rm{T})
\end{array}

$ \text{Step 7: Verification} $

\begin{array}{c}
\text{Moment Check: } \sum M_H = 0 \quad \checkmark \\
\text{Force Balance: } \sum F_x = \sum F_y = 0 \quad \checkmark \\
\hline
\text{Final Results:} \\
\text{CD: } 3375\,{\rm{lb}}\,(\rm{C}) \\
\text{CH: } 5625\,{\rm{lb}}\,(\rm{C}) \\
\text{HI: } 6750\,{\rm{lb}}\,(\rm{T})
\end{array}

$ \text{Step 8: Conclusion} $

\begin{array}{c}
\text{The truss maintains equilibrium with:} \\
\text{• CD resisting $3375\,{\rm{lb}}$ compression} \\
\text{• CH carrying $5625\,{\rm{lb}}$ compression} \\
\text{• HI sustaining$ 6750\,{\rm{lb}}$ tension}
\end{array}