The voltage at Tr3's base does change because the current through the emitter resistor is not constant. As I said before, adding a constant current sink makes it easier to understand. I admit I didn't understand this circuit until I analysed its behaviour with a constant current sink. A constant current has a near infinite impedance.
The LED drops 1.8V, V
BE is 0.6V so the voltage across the 1k resistor is around 1.2V, making the current equal to 1.2mA, irrespective of the collector voltage.
If decent/modern transistors are used (Hfe >200) the tiny current used to bias the bases can be ignored for the purposes of this exercise.
When the voltage at both inputs is equal, the current through both of the transistors and their 10k collector resistors will be equal. Half of the current will flow through one collector resistor and the other half through the other. In this case, the constant current sink takes 1.2mA so the current though the 10k collector resistors will be 0.6mA and the voltage across them 6V. Relative to the 0V point of the power supply, the output voltage will be equal to 9 - 6 = 3V.
As long as the voltage at both inputs is below 3V + V
BE = 3.6V and V
BE above the minimum voltage required for the current sink to regulate properly (the voltage across the 1k resistor, plus a little bit): -9 + 1.2V + 0.6V = -7.2V, then the output voltage will remain at 3V. This is known as the common mode range.