$ \text{Step 1: Given Data} $

The truss is subjected to:
${F_1} = 40\,{\rm{kN}}$ at E, ${F_2} = 50\,{\rm{kN}}$ at D,
${F_3} = 30\,{\rm{kN}}$ at C, ${F_4} = 40\,{\rm{kN}}$ at B.

Geometric properties:
Horizontal spacing $d = 2\,{\rm{m}}$, Vertical spacing $h = 1.5\,{\rm{m}}$.

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

\begin{array}{c}
\tan \theta = \frac{h}{d} = \frac{1.5}{2} = 0.75 \\
\theta = \tan^{-1}(0.75) = 36.87^\circ
\end{array}

$ \text{Step 3: Support Reaction at A} $

\begin{array}{c}
\sum M_A = 0: \\
F_4(h) + F_3(2h) + F_1(d) – F_y(2d) = 0 \\
(40)(1.5) + (30)(3) + (40)(2) – F_y(4) = 0 \\
\hline
F_y = 57.5\,{\rm{kN}}\,\left( \uparrow \right)
\end{array}

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

\begin{array}{c}
\sum M_E = 0: \\
-F_y(d) – F_{GH}(h) = 0 \\
-(57.5)(2) – F_{GH}(1.5) = 0 \\
\hline
F_{GH} = -76.67\,{\rm{kN}}\,\left( {\rm{T}} \right)
\end{array}

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

\begin{array}{c}
\sum F_y = 0\,\text{at E}: \\
F_y + F_{EH}\sin\theta – F_1 = 0 \\
57.5 + F_{EH}(0.6) – 40 = 0 \\
\hline
F_{EH} = -29.2\,{\rm{kN}}\,\left( {\rm{T}} \right)
\end{array}

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

\begin{array}{c}
\sum F_x = 0\,\text{at E}: \\
-F_{GH} – F_{ED} – F_{EH}\cos\theta = 0 \\
-(-76.67) – F_{ED} – (-29.2)(0.8) = 0 \\
\hline
F_{ED} = 100\,{\rm{kN}}\,\left( {\rm{C}} \right)
\end{array}

$ \text{Step 7: Results Summary} $

\begin{array}{c}
\text{ED} = 100\,{\rm{kN}}\,\left( {\rm{C}} \right) \\
\text{EH} = 29.2\,{\rm{kN}}\,\left( {\rm{T}} \right) \\
\text{GH} = 76.67\,{\rm{kN}}\,\left( {\rm{T}} \right)
\end{array}