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Given that the angle between a tangent to a circle and a chord drawn from the point of contact is equal to an angle in the alternate segment and tangent $T A N$ to circle $O$, and chord $A C$ subtending $\hat{B}$. Prove that $\hat{A}_{1}=\hat{C}_{2}$

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PROOF:

Draw in diameter $A O D$ an join $D C$.
\begin{aligned} \hat{A_{1}}+\hat{A_{2}} &\left.=90^{\circ} \text { ( } \tan \perp \text { radius }\right) \\ \hat{C}_{1}+\hat{C}_{2} &=90^{\circ}(\angle \text { in semi-circle }) \\ \hat{A}_{2} &=\hat{C}_{1}(\angle \text { 's in same seg }) \\ \therefore \hat{A}_{1} &=\hat{C}_{2} \end{aligned}

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