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!*)