From the FBD of member BC

$\Sigma F_V = 0$
$C_V = B_V = \frac{1}{2}(20)$

$C_V = B_V = 10 ~ \text{kN}$

$\Sigma F_H = 0$

$C_H = B_H$

From the FBD of Member AB

$\Sigma M_A = 0$
$4B_H = 3(10) + 4(10)$

$B_H = 17.5 ~ \text{kN}$

$\therefore C_H = 17.5 ~ \text{kN}$

$\Sigma F_H = 0$

$A_H = B_H$

$A_H = 17.5 ~ \text{kN}$

$\Sigma F_V = 0$

$A_V = 10 + 10$

$A_V = 20 ~ \text{kN}$