In the given figure, $O$ is the center of a circle, $A B$ is a diameter and $A C$ is a chord. Prove that $A B > A C$ .

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Solution

$\begin{matrix} L e t O b e c e n t r e o f c i r c l e O A = O B = O C = R a d i u s o f c i r c l e O A + O C > A C O A + O B > A C A B > A C h e n c e, p r o v e d . \end{matrix}$

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