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  • In Figure-7, AB is the diameter of a circle with centre O and AC is its chord such that . If the tangent dr

In Figure-7, AB is the diameter of a circle with centre O and AC is its chord such that \angle BAC = 30^{\circ}. If the tangent drawn at C intersects extended AB at D, then show that BC = BD.

 

 
 
 
 
 

Answers (1)

\begin{aligned} &\ln \Delta A C B\\ &\angle CAB+\angleA CB+\angle CBA=180^{\circ}\\ &30^{\circ}+90^{\circ}+\angle \mathrm{CBA}=180^{\circ}\\ &\angle CBA=180^{\circ}-120^{\circ}\\ &\angle \mathrm{CBA}=60^{\circ} \ldots \ldots \ldots \ldots (i)\\ &\text { In } \Delta O C B \text { , }\\ &O C=O B \Rightarrow \angle \mathrm{OCB}=\angle \mathrm{OBC}\\ & \angle \mathrm{OCB}=60^{\circ} \end{aligned}

\\$ $\angle O C D=90^{\circ}$ (Perpendicular from center of tangent) $ \\ \angle \mathrm{OCB}+\angle \mathrm{BCD}=90^{\circ} \\$ $60^{\circ}+\angle \mathrm{BCD}=90^{\circ} \\$ $\angle B C D=30^{\circ}......(ii) \\$ $ \angle \mathrm{CBO}=\angle \mathrm{BCD}+\angle \mathrm{CDB}\\$ $60^{\circ}=30^{\circ}+\angle \mathrm{CDB}\\$ $\angle \mathrm{CDB}=30^{\circ} .....(iii) \\ $ From equation 2 and equation 3 $ \\ \angle \mathrm{CDB} = \angle \mathrm{BCD} \\ B C=B D$

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