From the given circuit

\begin{gathered} \frac{V_{ A }-20}{10}+\frac{V_{ A }}{10}+\frac{V_{ A }}{20}=0 \\ 2  V_{ A }-40+2  V_{ A }+V_{ A }=0 \\ 5 V_{ A }=40 \Rightarrow V_{ A }=8  V \\ I=\frac{8}{20}=0.4  A \\ V_{ CD }=10 \times 0.4=4  V \end{gathered}

The equivalent resistance across the terminals, when all active sources are zero is

\left\lgroup \frac{10 \times 10}{20}+10 \right\rgroup \| 10=\frac{15 \times 10}{25}=6  \Omega

So, the Thevenin equivalent circuit is