ac voltage added to dc base line
Your second schematic bears no relation to the first.  The voltage at the nexus of the two resistors and one capacitor will be the sum of the voltages at the other ends of those components in inverse proportion to their impedances.  

In other words, the two resistors and one capacitor form a three-way impedance divider (ignoring the transistor for now) which is like a resistive divider except that the "resistance" depends on frequency.  You can think of an ideal capacitor as a resistor that has infinite resistance at DC and decreasing impedance as frequency increases.  So we have three inputs: 


0V (ground) via 100R
1V (V2) via 100R
a 1kHz sine wave via some capacitance C1


If we disconnect the sine wave source, we have a simple resistance divider, and since we have equal resistors, the output of the divider will be 0.5V.  The value of the capacitor is not given, but if it's 2.2uF, then it will also be about 100R at 1kHz, in which case the output of the impedance divider will be the average of the three inputs.  At lower frequencies, the impedance of the capacitor is higher, and so V1 will have less impact on the output of the divider.  At higher frequencies, the impedance of the capacitor is lower, and thus V1 will have a larger impact on the resulting voltage.

All of that said, we've been ignoring the transistor in the circuit.  The base-emitter junction of a transistor acts something like a diode with a nonlinear relationship between current and voltage, so where the voltage at the output of the impedance divider would increase, this is tempered by the fact that current will be driven into the base of the transistor.  Likewise, where the voltage would decrease, less current is driven into the transistor base.  This affects the actual voltage at the node where the two resistors, capacitor, and transistor base all meet up, but the details depend somewhat on how the other two terminals of the transistor are connected.