Derivation of Transfer Coefficient for Barrier Potential $E > V_o$

Given the following:

The potential is:

$$V(x) = 0$$ , $$x<0$$

$$V(x) = V_o$$ , $$0\leq x\leq a$$

$$V(x) = 0$$ , $$x > a$$

In the three regions the solutions to Schrondinger Eq. are:

$$\psi_1 = Ae^{ik_1x} + Be^{-ik_1x}$$

$$\psi_2 = Ce^{ik_2x} + De^{-ik_2x}$$

$$\psi_3 = Fe^{ik_3x} + Ge^{-k_3x}$$

where

$$k_1 = k_3 = \sqrt{\frac{2mE}{\hbar^2}}$$

$$k_2 = \sqrt{\frac{2m(E-V_o)}{\hbar^2}}$$

The goal is to show that

$$T = \frac{j_{transmitted}}{j_{incident}}= \frac{|F|^2}{|A|^2}$$

reduces to

$$T = \frac{1}{1 + \frac{1}{4}\frac{V_o^2}{E(E-V_o)}sin^2k_2a}$$ (Eq. 2.37 in the textbook)

I am unable to make any progress after the below steps

$$A + B = C + D = F + G$$

$$A + B = C + D = F + 0$$

$$k_1(A - B) = k_2(C - D) = k_3F$$

$$F = A - B = \frac{k_2(C - D)}{k_3}$$

$$A = \frac{k_2(C - D)}{k_3} + B = F + B$$

$$T = \frac{(\frac{\sqrt{\frac{2m(E-V_o)}{\hbar^2}}(C-D)}{\sqrt{\frac{2mE}{\hbar^2}}})^2}{(\frac{\sqrt{\frac{2m(E-V_o)}{\hbar^2}}(C-D)}{\sqrt{\frac{2mE}{\hbar^2}}}+B)^2}$$

How can I get the above into the desired form of Eq. 2.37?

Remember that for the wavefunction, we have two sorts of boundary conditions: continuity, AND continuity of its derivatives (where the latter applies if we do not have an infinite potential). So when you write the first equation comes from continuity of $$\psi$$:
$$\psi_1(x=0) = \psi_2(x=0) \implies A + B = C + D$$
$$\psi_1'(x=0) = \psi_2'(x=0)$$
and there will also be similar equations for $$\psi_2$$ and $$\psi_3$$. (And since this is a homework question, you have the fun of working it out!)