Question Number 44636 by Raj Singh last updated on 02/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18
$${join}\:{oB}\:{andOC} \\ $$$$\bigtriangleup{OAB}\:{and}\:{OAC}\: \\ $$$$\left.\mathrm{1}\right){OB}={OC}\:\left({radius}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left.\mathrm{2}\right)\angle{ABO}=\angle{ACO}=\mathrm{90}^{{o}} \\ $$$$\left.\mathrm{3}\right){OA}\:{common} \\ $$$${so}\:\:\bigtriangleup{OAB}\:{congurent}\:\bigtriangleup{OAC} \\ $$$${so} \\ $$$$\angle{OAB}=\angle{OAC} \\ $$$${AB}={AC} \\ $$$${let}\:{BC}\:{interzect}\:{OA}\:{at}\:{point}\:{P} \\ $$$$\:\:{in}\:\:\bigtriangleup{APB}\:{and}\:\bigtriangleup{APC} \\ $$$${AB}={AC} \\ $$$${AP}\:{common} \\ $$$$\angle{OAB}=\angle{OAC} \\ $$$${so}\:\bigtriangleup{APB}\:{congurent}\:\bigtriangleup{APC} \\ $$$${so}\:\:\angle{APB}=\angle{APC}=\alpha \\ $$$$\alpha+\alpha=\mathrm{180}^{{o}} \\ $$$$\alpha=\mathrm{90}^{{o}} \:{hence}\:{BC}\bot{OA} \\ $$$${BP}={PC} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$