This question was inspired by incircle bisectors, but in this case two inscribed equilateral triangles, oriented along the corresponding angle bisector, play the role of the two incircles:
Using the known expression for the length of bisector \begin{align} |CC_0|&=\frac{2 ab\cos\tfrac\gamma2}{a+b} , \end{align} it is easy to find the side length of the inscribed $\triangle C_0C_1C_2$:
\begin{align} |C_0C_1|=|C_1C_2|=|C_0C_2|= q &= \frac{4ab}{(a+b) \Big(1+\sqrt3\tan\tfrac\gamma2\Big) \,\sqrt{1+\cot^2\tfrac\gamma2}} . \end{align}
Now, consider the bisection of $\triangle ABC$ with two inscribed equilateral triangles
Let \begin{align} |AC|&=b ,\quad |BC|=a ,\quad |AB|=c ,\\ |AD|&=c_1 ,\quad |BD|=c_2=c-c_1 ,\\ |CD|&=d ,\quad |CC_{00}|=d_1 ,\quad |CC_{10}|=d_2 ,\\ |C_{00}C_{01}|&=|C_{01}C_{02}|=|C_{00}C_{02}| =q_1 ,\\ |C_{10}C_{11}|&=|C_{11}C_{12}|=|C_{10}C_{12}| =q_2 ,\\ \angle CAB&=\alpha ,\quad \angle ABC=\beta ,\quad \angle BCA=\gamma ,\\ \angle DCA&=\gamma_1 ,\quad \angle BCD=\gamma_2=\gamma-\gamma_1 ,\\ \angle ADC&=\delta_c . \end{align}
For this configuration we have
\begin{align} q_1&= \frac{4bd}{(b+d)(1+\sqrt3\tan\tfrac{\gamma_1}2)\,\sqrt{1+\cot^2\tfrac{\gamma_1}2}} \tag{1}\label{1} ,\\ q_2&= \frac{4ad}{(d+a)(1+\sqrt3\tan\tfrac{\gamma_2}2)\,\sqrt{1+\cot^2\tfrac{\gamma_2}2}} \tag{2}\label{2} . \end{align}
The question is: solve \eqref{1}-\eqref{2} for $q_1=q_2$.
It is relatively easy to express \eqref{1}-\eqref{2} in terms of single parameter, say, $c_1$ and solve this numerically, but is there a way to find simpler analytic solution e.q. in a polynomial form?