Consider a triangle ABC in the plane with angles $A, B, C$ and side lengths $a, b, c$ with the side of length $a$ opposite the angle $A$, etc, as usual.
Suppose also that the angles appear in the order $A, B, C$ as one traverses the vertices in an anticlockwise direction. Regard the three vertices as complex numbers $\alpha, \beta, \gamma$ with $\alpha$ corresponding to the vertex with angle $A$, etc.
So a triangle corresponds to a point $(\alpha, \beta, \gamma)$ in $3$-dimensional complex space. Not all points in $3$-dimensional complex space correspond to triangles since the points defined may be collinear.
I need to show that for the described triangle the following equation holds:
$$(be^{iA} − c)\alpha − be^{iA}\beta + c\gamma = 0$$
I'm pretty stumped! Any help would be appreciated.
Given a triangle constructed as described in the question, the complex number $\beta - \alpha$ corresponds to the side of the triangle opposite the vertex $\gamma,$ so it has modulus $c.$ The complex number $\gamma - \alpha$ corresponds to the side of the triangle opposite the vertex $\beta,$ so it has modulus $b.$ The complex number $\gamma - \alpha$ also has argument $\arg(\gamma - \alpha) = \arg(\beta - \alpha) + A,$ because the magnitude of the angle between the two sides represented by $\beta - \alpha$ and $\gamma - \alpha$ is $A$ and because $\gamma$ is anticlockwise from $\beta$ when viewed from the position of $\alpha.$
In other words, if $\psi = \arg(\beta - \alpha),$ then \begin{align} \beta - \alpha &= c e^{i\psi}, \\ \gamma - \alpha &= b e^{i(\psi + A)} = be^{iA} e^{i\psi}. \end{align}
Now evaluate $c(\gamma - \alpha) - be^{iA}(\beta - \alpha)$ and compare the result to the fact that is to be proved.