From the FBD of the pulley
$\Sigma F_x = 0$
$A_H = 60 ~ \text{kN}$
$\Sigma F_y = 0$
$A_V = 60 ~ \text{kN}$
From the FBD of the whole frame
$\Sigma M_E = 0$
$4C_V = 6(60)$
$C_V = 90 ~ \text{kN}$
$\Sigma M_C = 0$
$4E_V = 10(60)$
$E_V = 150 ~ \text{kN}$
From the FBD of member AC
$\Sigma F_y = 0$
$B_V = 60 + 90$
$B_V = 150 ~ \text{kN upward}$ answer
$\Sigma M_C = 0$
$3B_H + 8(60) = 4B_V + 6(60)$
$3B_H + 480 = 4(150) + 360$
$B_H = 160 ~ \text{kN to the left}$ answer
Check Through Member DE
$\Sigma M_E = 0$
$3B_H = 8(60)$
$B_H = 160 ~ \text{kN}$ (okay!)
$\Sigma F_y = 0$
$B_V = 150 ~ \text{kN}$ (okay!)