Let $ABC$ be a triangle $A \neq B$ and let $P \in (AB)$ be a point for which denote $m(\widehat{ACP}) = x$ and $m(\widehat{BCP}) = y$. Prove that $\frac{\sin A \sin B}{\sin(A-B)} = \frac{\sin x \sin y}{\sin(x-y)}$ if and only if $PA = PB$.
Suppose WLOG $A > B$ and hence $x > y$. Denote the circumcentre $w$ of the $\triangle ABC$ and the second point $M \in (P \cap W)$. Construct the points $X \in W, Y \in W$ for which $MX \| AB \| CY$. Observe that $m(\widehat{CMY}) = A - B$. $$m(\widehat{XYM}) = x - y \text{ and } PC/PM = CB \cdot CA / M_A \cdot MB = \frac{\sin A \sin B}{\sin x \sin y}$$ Observe that, $$CY/MX = \sin(A-B)/\sin(x-y)$$ hence $\frac{\sin A \sin B}{\sin(A-B)} = \frac{\sin x \sin y}{\sin(x-y)} \Rightarrow PC/PM = CY/MX \Rightarrow PEXY \Rightarrow PA = PB$
Let $ABC$ be a triangle $A \neq B$ and let $P \in (AB)$ be a point for which denote $m(\widehat{ACP}) = x$ and $m(\widehat{BCP}) = y$. Prove that $\frac{\sin A \sin B}{\sin(A-B)} = \frac{\sin x \sin y}{\sin(x-y)}$ if and only if $PA = PB$.