Let force is each member be denoted as member's name. We'll take support reactions at A as $H_A$ & $V_A$ and support reaction at roller E as $R_E $

Applying COE to whole truss,

$\sum F_x = 1.5 + H_A = 0 \space\space\space\space \sum F_Y = -3 + R_E + V_A = 0 \\ H_A = -1.5 KN =1.5 KN \space\space R_E + V_A = 0 $

Now,

$\sum MFA\cong 0 [ +ve] \\ 1(4) + 1(8) + 1(12) + 1.5(4) - RE(8) = 0 \\ R_E = 3.75KN \\ So, VA =3.75KN $

Now, for finding force in members BD, BE, CE cutting members BD, BE, CE using sectional plane 1-1

Considering r.h.s. part

$\theta=\tan^{-1}(2/4) = 26.565^\circ $

Now, Applying COE,

$\sum F_x= -BD \cos 26.56 - BE \cos 26.56 - CE + 1.5 = 0 \\ \sum F_Y=R_E - BD \sin 26.565 + BE \sin 26.56 - 1- 1= 0 \\ \sum MFE\cong 0 [ +ve] \\ -BD (\cos 26.56)(4) + 4(1) + 1.5(4)=0 \\ BD = 2.795 KN (T) $

Putting value in $\sum F_Y$ equation

$$BE = -1.12 = 1.12 (C) $$

Putting in $\sum F_x$ equation, we get $CE=0 $

Now, applying method of joints to find forces in other members (considering all unknown forces to be tensile)

Consider joint A

$\sum F_x = H_A + AC + AB \cos \theta = 0 \\ \sum F_Y = AB \sin \theta + V_A = 0 $

Solving we get,

$AB = 1.68 (T) $ & $AC =0$

Consider joint D

$\sum F_Y = - 1 - DE - BD \sin 26.56 = 0 \\ DE= -2.25 = 2.25(C) \\ \sum F_X = - BD \cos 26.56 + DF = 0 \\ DF = 2.5 (T)$

Consider joint F

$\sum F_Y = -FE \sin 45 - 1 = 0 \\ FE = -1.414 = 1.414(C) $

Also, at joint C, as 2 members are collinear, force in third member BC will be 0

Hence, forces in all members can be summarised as below.