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$\Huge\textsf{圆中角平分线的一个小结论}$

如图，$A,B,C$为圆上三点，$AD$平分$\angle BAC,$用$AB ,AC ,\alpha$表示$AD$.

解：延长$AB$至$F$，使得$BF = CA.$

$\because \angle BAD = \angle CAD ,\;\therefore BD = CD.$

$\because$四边形$ABDC$为圆内接四边形，$\therefore \angle FDB = \angle ACD.$

又$\because BF = CA,\;\therefore \triangle FBD \cong \triangle ACD \;(SAS)$

$\therefore DF = DA.$

作$DE \perp AF.$

$\because DA = DF ,\;\therefore AE = BE = \dfrac{1}{2}AF = \dfrac{1}{2}(AB + AC).$

$\therefore AD = \dfrac{AE}{\cos \alpha}= \dfrac{AB+AC}{2\cos \alpha}.$

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