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Question Number 175149 by cherokeesay last updated on 20/Aug/22

Answered by som(math1967) last updated on 21/Aug/22

Commented by som(math1967) last updated on 21/Aug/22

$$AB=2R,BC=R$$$$\therefore AC=\sqrt{4R^2+R^2}=\sqrt{5}R$$$$AE=AB-BE=2R-r$$$$△\mathit{A}\mathit{B}\mathit{C}\sim △\mathit{A}\mathit{D}\mathit{E}$$$$\therefore \frac{2R-r}{\sqrt{5}R}=\frac{r}{R}$$$$\Rightarrow 2R-r=\sqrt{5}r$$$$\Rightarrow 2R=(\sqrt{5}+1)r$$$$\therefore \frac{R}{r}=\frac{\sqrt{5}+1}{2}=\mathrm{\Phi }$$

Commented by cherokeesay last updated on 21/Aug/22

$$Nice$$$$thankyou!$$

Commented by Tawa11 last updated on 21/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

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