Isosceles triangle

We do not loose generality if we place the triangle with the base in $(-1,0),\;(1,0)$.

Let's call the angles $\angle1, \, \cdots, \, \angle 4$ as $\alpha_1,\, \cdots, \, \alpha_4$ (just to better handle them symbolically), then we must have $$ \bbox[lightyellow] { \alpha _{\,1} + \alpha _{\,2} = \alpha _{\,3} + \alpha _{\,4} \quad \Rightarrow \quad \left\{ \matrix{ \alpha _{\,4} - \alpha _{\,1} = \alpha _{\,2} - \alpha _{\,3} = \delta \hfill \cr \alpha _{\,4} + \alpha _{\,1} = \delta + 2\alpha _{\,1} = \delta + \beta \hfill \cr \alpha _{\,2} + \alpha _{\,3} = \delta + 2\alpha _{\,3} = \delta + \gamma \hfill \cr} \right. } \tag{1}$$

Consider now two segments of length $r$ departing from points $B$ and $C$, and ending at points $$ \bbox[lightyellow] { \eqalign{ & D = \left( { - 1 + r\cos \left( {\delta + \beta /2} \right),\;r\sin \left( {\delta + \beta /2} \right)} \right) \cr & E = \left( {1 - r\cos \left( {\delta + \gamma /2} \right),\;r\sin \left( {\delta + \gamma /2} \right)} \right) \cr} } \tag{2}$$ so that they satisfy the conditions imposed on them for the length, and for the angles (they shall bisect $\beta$ and $\gamma$).

Let's consider then the lines $C,D$ and $B,E$. Their equations are $$ \bbox[lightyellow] { \left\{ \matrix{ {\rm line}\,{\rm CD}:\;{{x - 1} \over { - 2 + r\cos \left( {\delta + \beta /2} \right)}} = {y \over {r\sin \left( {\delta + \beta /2} \right)}} \hfill \cr {\rm line}\,{\rm BE}:{{x + 1} \over {2 - r\cos \left( {\delta + \gamma /2} \right)}} = {y \over {r\sin \left( {\delta + \gamma /2} \right)}} \hfill \cr} \right. } \tag{3}$$ and we want their slopes to be: $$ \bbox[lightyellow] { \eqalign{ & \left\{ \matrix{ {{r\sin \left( {\delta + \beta /2} \right)} \over { - 2 + r\cos \left( {\delta + \beta /2} \right)}} = - \tan \left( {\delta + \gamma } \right) \hfill \cr {{r\sin \left( {\delta + \gamma /2} \right)} \over {2 - r\cos \left( {\delta + \gamma /2} \right)}} = \tan \left( {\delta + \beta } \right) \hfill \cr} \right.\quad \Rightarrow \quad (4.a) \cr & \Rightarrow \quad \left\{ \matrix{ {{\sin \left( {\delta + \beta /2} \right)} \over {2/r - \cos \left( {\delta + \beta /2} \right)}} = \tan \left( {\delta + \gamma } \right) \hfill \cr {{\sin \left( {\delta + \gamma /2} \right)} \over {2/r - \cos \left( {\delta + \gamma /2} \right)}} = \tan \left( {\delta + \beta } \right) \hfill \cr} \right.\quad \Rightarrow \quad (4.b) \cr & \Rightarrow \quad \left\{ \matrix{ \sin \left( {2\delta + \beta /2 + \gamma } \right) = 2/r\sin \left( {\delta + \gamma } \right) \hfill \cr \sin \left( {2\delta + \beta + \gamma /2} \right) = 2/r\sin \left( {\delta + \beta } \right) \hfill \cr} \right. \quad (4.c) \cr} }$$

The system of equations in (4.c) above can be represented as $$ \bbox[lightyellow] { \left\{ \matrix{ 0 \le \beta ,\gamma < \pi /2 - \delta \hfill \cr F\left( {\beta ,\;\,\gamma \;;\;\,\delta ,r} \right) = \sin \left( {2\delta + \beta /2 + \gamma } \right) - 2/r\sin \left( {\delta + \gamma } \right) \hfill \cr F\left( {\beta ,\;\,\gamma \;;\;\,\delta ,r} \right) = 0 \hfill \cr F\left( {\gamma ,\;\beta \;\,;\,\;\delta ,r} \right) = 0 \hfill \cr} \right. } \tag{5}$$ and since it imposes to be null either the $F(\beta,\,\gamma)$ and its symmetric $F(\gamma,\, \beta)$, then, clearly, if there are, the solutions will be $\beta=\gamma$, i.e. the triangle must be isosceles.
Q.E.D.