similarity of triangles in the complex plane

Question

The problem reads: given a triangle with vertices $\alpha, \beta,\gamma$ in the complex plane labeled anti-clockwise. Denote the angle at the vertex $\alpha$ by $A$, the one at $\beta$ by $B$ and the one at $\gamma$ by $C$. Show that any triangle in the complex plane with side lengths $a,b,c$ is similar to triangle $(\alpha,\beta,\gamma)$ (the sides are labeled so that $a$ is opposite the angle $A$, $b$ opposite angle $B$ etc. ) iff the following holds: $$ a^2\alpha^2 + b^2\beta^2 + c^2\gamma^2 + (a^2-b^2-c^2)\beta\gamma + (b^2-c^2-a^2)\alpha\gamma+ (c^2-a^2-b^2)\alpha\beta =0 $$ I tried to show that if they are similar then the relation holds, I derived 3 equations : $$ e^{iA} =\frac{c(\gamma-\alpha)}{b(\beta-\alpha)}\\ e^{iB} = \frac{a(\alpha-\beta)}{c(\gamma-\beta)}\\ e^{iC} = \frac{b(\beta-\gamma)}{a(\alpha - \gamma)} $$ (note that I assumed the sides of the 2 triangles to be at constant ratios since they are similar) from these I can use the cosine rule on the triangle with sides $a,b,c$ and get: $$ \frac{a^2-b^2-c^2}{bc} = \frac{c(\gamma-\alpha)}{b(\beta-\alpha)} + \frac{b(\beta-\alpha)}{c(\gamma-\alpha)}\\ \frac{b^2-a^2-c^2}{ac} = \frac{a(\alpha-\beta)}{c(\gamma-\beta)} + \frac{c(\gamma-\beta)}{a(\alpha-\beta)}\\ \frac{c^2-a^2-b^2}{ba} = \frac{b(\beta-\gamma)}{a(\alpha - \gamma)} + \frac{a(\alpha - \gamma)}{b(\beta-\gamma)} $$ I expanded them many times and I can never get the answer. Does anybody have a different approach?

Answer 1

Ok sorry everybody. The signs on RHS of the equations I derived are wrong. They should all be negative. Making this correction yields the right result after expanding brackets and adding the 3 equations.