Thevenin theorem Hi i got the trouble and i miss the concept about v out in thevenin 
i have the circuit like this 

so i used voltage divider for finding the voltage at point 1,
 but how to finding voltage at Vo , is that same the result of voltage at 2 , and voltage between R2 and R3 , 
i thought at R2 and R3 we need voltage divider calculation also , from voltage at point 2, is that true? . if The calculation like that there is a voltage difference between two Vout
 A: It sounds as though you're on the right track: if I understand correctly, you're saying that once you have found the voltage across the pair $R_1 and R_4$ (equal to the voltage across $R_2$ and $R_3$), presumably by lumping $R_1,\,R_2\,R_3\,R_4$ together through parallel addition of $R_1 +R_4$ and $R_2+R_3$, then you simply work out $V_o$ thinking of $R_2,\,R_3$ as a voltage divider. This is indeed correct. 
However I am a little puzzled by your mention of Thévenin if the above is your approach. Presumably your question asks for you to use Thévenin's theorem, which is simply a big name for the statement that the output current to voltage relationship is linear, and you can fully define such a relationship by drawing a straight line between two points, which happen in Thévenin's "theorem" to be the voltage at zero current (open circuit) and current at zero voltage (dead short). 
So, take out $R_1$ and calculate the voltage across the gap: this is simply the voltage across $R_1$ and $R_4$ with $R_2$ and $R_3$ removed. This is your open circuit voltage $V_T$. Now consider a dead short in place of $R_2$ and calculate the current $I_T$ through $R_3$ in this configuration. By linearity, the circuit is equivalent to $V_T$ driving $R_2$ through the Thévenin resistance $V_T/I_T$. Now you apply the voltage divider notion  to the ladder $R_2$ and $R_T$ to find $V_o=E_1\,R_2/(R_T+R_2)$.
In this special case, there is an easy approach by symmetry. By the mirror symmetry, the circuit is equivalent to $E_1/2=6V$ driving $R_5$ in series with $R_1$ paralleled with $R_2$. This means that the $R_1$ paralled with $R_2$ (with half the resistance of $R_5$) is going to feel a third of the source's voltage across it, namely $2V$. 
