Thevenin Equivalent does not seem equivalent for me
In the equivalent circuit, with no load there is no voltage drop across the 25 ohm resistor (since there is no current through it), so the output is 50 volts.

Now let's say you add a 50 ohm load. The top circuit  has a lower branch equivalent to 25 ohms (two 50's in parallel, and will give an output of $$ V = 100 (\frac{25}{25 + 50}) = 33.3 \text{ volts} $$ The equivalent has 50 ohms between A and B, so $$ V = 50(\frac{50}{25 + 50}) = 33.3\text{ volts}$$ So they really are equivalent.