I'm trying to show that a triangulation $\tau_h$ is regular if and only if there exists $\theta_0>0$ such that for all $T\in\tau_h$ we have $\theta_T\geq \theta_0>0$, whereas $\theta_T$ is the smallest angle of the triangle $T$.
I was able to prove the "$\Leftarrow$" part of this:
Since $\tau_h$ is shape regular we have $C>0$ such that for all $T\in\tau_h$ we have $\rho_T/h_T \geq C>0$ whereas $\rho_T$ is the radius of the in-circle of $T$ and $h_T$ the radius of the out-circle of $T$. Using some trigonometric relations we find $$\rho_T=4h_T\sin\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\sin\left(\frac{\gamma}{2}\right)$$
and so we have $\rho_T/h_T = 4\sin\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\sin\left(\frac{\gamma}{2}\right)$. Since $\alpha/2, \beta/2, \gamma/2$ are all smaller than 90° we can estimate this by
$\rho_T/h_T \geq 4\sin^3\left(\frac{\theta_0}{2}\right)=:C>0$ and hence we have shown the "$\Leftarrow$"-part.
What I haven't been able to prove was the other direction. Assuming I have some constant $C>0$ such that $\frac{\rho_T}{h_T}\geq C > 0$, how can I derive the existence of a lower boundary $\theta_0>0$ for the smallest angle ?
You can argue by contradiction.
Without loss of generality, assume $\alpha < \theta_0$. By the identity you gave: $$ \frac{\rho_T}{h_T}=4\sin\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\sin\left(\frac{\gamma}{2}\right) < 4\frac{\alpha}{2}<2\theta_0 , $$ for $0<\beta/2<\pi/2$ and $0<\gamma/2<\pi/2$. Choosing $\theta_0< C/2$, we have $\rho_T/h_T < C$. Contradiction.
Therefore if $\dfrac{\rho_T}{h_T}\geq C$ for some $C>0$, there exists a $\theta_0$, such that the minimum angle is greater than or equal to $\theta_0$.