From the FBD of the whole system
$\Sigma M_D = 0$
$2A_x = 3(60) + 6(30)$
$A_x = 180 \, \text{ kN}$
$\Sigma M_A = 0$
$2D_x = 3(60) + 6(30)$
$D_x = 180 \, \text{ kN}$
From the FBD of member ABC
$\Sigma F_x = 0$
$\frac{3}{5}F_{CF} = A_x$
$\frac{3}{5}F_{CF} = 180$
$F_{CF} = 300 \, \text{ kN}$
$\Sigma M_B = 0$
$3A_y = 1.5(\frac{4}{5}F_{CF})$
$3A_y = 1.2(300)$
$A_y = 120 \, \text{ kN}$
From FBD of the whole system
$\Sigma F_V = 0$
$D_y = A_y + 60 + 30$
$D_y = 120 + 60 + 30$
$D_y = 210 \, \text{ kN}$
Summary
$A_x = 180 \, \text{ kN leftward}$
$A_y = 120 \, \text{ kN downward}$
$D_x = 180 \, \text{ kN rightward}$
$D_y = 210 \, \text{ kN upward}$
$F_{CF} = 300 \, \text{ kN tension}$
