In the following diagram, $E$ is a point inside triangle $ABC$, and $\overline{AE}, \overline{BE}, \overline{CE}$ are drawn in. Angles $\alpha_1, \alpha_2, \beta_1, \beta_2, \gamma_1, \gamma_2$ are labeled as shown.
Considering either the sum of angles of $\triangle ABC$ or the sum of angles around $E$, the angles must satisfy $$ \alpha_1 + \alpha_2 + \beta_1 + \beta_2 + \gamma_1 + \gamma_2 = \pi. \tag{1} $$
But there must be another condition. Indeed, if $\alpha_1, \alpha_2, \beta_1$, and $\beta_2$ are fixed, then $\gamma_1$ and $\gamma_2$ are determined.
So my question is: Given (1), what is the additional relation between the angles $\boldsymbol{\alpha_1}$, $\boldsymbol{\alpha_2}$, $\boldsymbol{\beta_1}$, $\boldsymbol{\beta_2}$, $\boldsymbol{\gamma_1}$, $\boldsymbol{\gamma_2}$?
Somehow I was under the impression that angle chasing suffices to obtain all of the angle dependencies in a diagram. Evidently I am wrong, as angle chasing in the above diagram does not appear to yield anything other than (1).
By the law of sines:
$$ \frac{CE}{BE} = \frac{\sin \alpha_1}{\sin \alpha_2}, \;\;\; \frac{AE}{CE} = \frac{\sin \beta_1}{\sin \beta_2}, \;\;\; \frac{BE}{AE} = \frac{\sin \gamma_1}{\sin \gamma_2} $$
Multiplying together:
$$ \sin \alpha_1\;\sin\beta_1\;\sin \gamma_1 \;=\; \sin \alpha_2\;\sin\beta_2\;\sin \gamma_2 $$