Let force is each member be denoted as member's name

We'll take support reactions at A as HA & VA

And support reaction at roller D as RD

$Also, CD = 6 \tan30 = 3.464m \\ AD = 6/\cos 30 = 6.93m $

Applying COE to whole truss,

$\sum Fx = HA + 20 \cos60 + 40 \cos60 + 20 \cos60 = 0 \\ HA= - 40 KN = 40 KN \\ \sum FY = VA + RD - 20 \sin60 - 40 \sin60 - 20 \sin60 = 0 \text{____} (i) \\ \sum MFA= 0 [ +ve] \\ 40(3) + 20(6) - RD(6.93) = 0 \text {___}(ii) \\ RD = 34.63 KN $

So, from (i) &(ii)

$VA = 34.652 KN $

Now, cutting members BC, CE & ED by cutting plane 1-1

Assuming all of them to be tensile

Applying COE

$\sum Fx = - BC \cos30 - CE \cos60 - ED + 20 \cos60 = 0 ------(i) \\ \sum FY = - 20 \sin60 - BC \sin30 - CE \sin60 + RD = 0 --------(ii) \\ \sum MFC= 0 [ +ve] \\ 3*ED - 1.732(RD) = 0 \\ ED = 20 KN (T) ----(iii) $

From (i)(ii) &(iii),we get

$BC = -34.63 KN = 34.63 KN (C) \\ CE = 40 KN (T) $

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

Considering joint C,

$\sum Fx = 20 \cos60 - BC \cos30 - CE \cos60 + CD \sin 30 =0 \\ CD = - 20 KN = 20 KN (C) $

Considering joint B

These 4 forces are forming 2 pairs of collinear forces therefore, $BC = AB = -34.63 KN $ and $BE = -40 KN $

Considering joint A
$\sum Fx = AB \cos30 + AE + HA + 20 \cos30 = 0 \\ \sum FY = VA - 20 \sin60 + AB \sin 30 = 0 \\ AB = - 49.3 KN \\ \text {Hence, AE } = 72.695 KN $

Hence, forces in all members can be summarised as below