Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(c,0)$. System of the three perpendicular bisectors $\left\{\begin{array}{lll} x=\frac{c-b}{2} \\ y-\frac{a}{2}=-\frac{b}{a}(x+\frac{b}{2}) \\ y-\frac{a}{2}=\frac{c}{a}(x-\frac{c}{2} \end{array}\right.$ Solution $O(\frac{c-b}{2},\frac{a^{2}-bc}{2a})$. Slope of the line $OD\ :\ m_{OD}=\frac{a^{2}-bc}{a(c-b)}=\frac{a}{b}=m_{AB}$, the slope of the line $AB$. $2a^{2}b=c(a^{2}+b^{2})$, $\frac{2ab}{a^{2}+b^{2}}=\frac{c}{a}$, $2 \cdot \frac{a}{\sqrt{a^{2}+b^{2}}} \cdot \frac{b}{\sqrt{a^{2}+b^{2}}}=\cot \gamma$, $2 \cdot \sin \beta \cdot \cos \beta=\cot \gamma$, $\sin 2\beta=\cot \gamma$.

Let $P$ be a point on $(ABC)$ such that $BP$ is the reflection of $AB$ w.r.t. $BC$. Then $\angle PBC=\angle ABC$ therefore $PC=AC$ then $CQ\perp AP$. Let $Q=CO\cap AP$ and $R=AO\cap (ACQ)$. $\angle ABP=\angle AOC= 2\widehat{B}$ then $\angle CAR=\angle CDR=90-\widehat{B}$ (1). We consider that $DO\parallel AB$ then $\angle ADO=\angle ADC-\angle ODC=90-\widehat{B}= (1)\rightarrow \Delta ADO\sim \Delta CDR$. Then $\frac{AD}{AO}=\frac{CD}{CR}\rightarrow \frac{CR}{AO}=\frac{CD}{AD}$ (2) We know that $AOC$ is isosceles, so $ACRQ$ is an isosceles trapezoid, therefore $CR=AQ$, now we rewrite (2) and consider that $\angle AOQ=180-2\widehat{B}$: $\frac{AQ}{AO}=\frac{CD}{AD}=\sin(180-2\widehat{B})=\cot(\widehat{C})$, then $\sin(2\widehat{B})=\cot( \widehat{C})$.

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