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Question Number 94925 by I want to learn more last updated on 21/May/20

Commented by mr W last updated on 22/May/20

$$AB=AC=BC=Radius$$$$\Rightarrow \mathrm{\Delta }ABC=equilateral$$$$\Rightarrow \mathrm{\angle }ABC=60°$$$$similarly$$$$\Rightarrow \mathrm{\angle }ABD=60°$$$$\Rightarrow \mathrm{\angle }CBD=2\times 60=120°$$

Commented by I want to learn more last updated on 22/May/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

Answered by ElOuafi last updated on 22/May/20

$$let\alpha =\mathrm{\angle }CBD,\beta =\mathrm{\angle }BCD=\mathrm{\angle }BDCand$$$$HtheverticaldroppingonCD$$$$BC=BD\Rightarrow △BCDisisosceleinB$$$$\Rightarrow 2\beta +\alpha =\pi$$$$inthetriangleABD$$$$AD=AB\wedge BD=AB\Rightarrow △ABDisequilateral$$$$\Rightarrow \mathrm{\angle }ABD=\mathrm{\angle }BDA=\mathrm{\angle }DAB=\frac{\pi }{3}$$$$\therefore inthetriangleBHD;(\mathrm{\angle }BHD=\frac{\pi }{2})$$$$\Rightarrow \beta +\frac{\pi }{3}+\frac{\pi }{2}=\pi \Rightarrow \beta =\frac{\pi }{6}$$$$\Rightarrow \alpha =\mathrm{\angle }CBD=\frac{2\pi }{3}$$

Commented by I want to learn more last updated on 22/May/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

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