Relationship between torque and back-EMF voltage | BLDC motor modeling

ea, eb, ec the back EMF voltages are proportional to the actual motor speed, unlike the applied voltages which are partiallly lost driving currents ia etc through the winding resistance. Thus, ea etc are the voltages generating useful mechanical power rather than heating the windings.

Then (ea.ia + eb.ib + ec.ic) is simply the electrical power which is converted to mechanical power P.

So the overall equation is simply

Torque = Power/rotational velocity.

At zero speed, both ea etc and omega are 0 so this form of the expression is indeterminate. Instead, you can use the speed constant Kv (rpm/volt), or its radians/sec/volt equivalent ... or indeed its inverse, which is the torque constant Kt, as you posit in the question.

If the $e_a$, $i_a$, etc., are the idealized voltages and currents on the armature, with magnetic and electrical losses neglected, then that equation simply follows from conservation of energy.

Let $P_e$ denote electrical power, and $P_s$ denote shaft power. Claim that there's no friction, resistance, or other losses. Then, in a universe that's not visibly changing dimensions, $P_e + P_s = 0$ by conservation of energy.

$$P_e = sum_{mathrm{all windings}}e_{winding}i_{winding}$$ $$P_s = T_e omega_m$$

Do an itty bitty bit of math, to substitute $e_a$, $i_a$, etc., for my pretentious summation over all hypothetical windings, then an itty bitty bit more by dividing both sides by $omega_m$, and you end up with your expression.