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Question Number 181852 by mnjuly1970 last updated on 01/Dec/22

Answered by mr W last updated on 01/Dec/22

$$\frac{a}{\sin \alpha }=\frac{b}{\sin \beta }=\frac{c}{\sin \gamma }=2R$$$$S_{ABC}=\frac{bc\sin \alpha }{2}=2R^2\sin \alpha \sin \beta \sin \gamma$$$$S_{ABC}=R(\sin \alpha +\sin \beta +\sin \gamma )r$$$$S_{SPQ}=2r^2\sin \mathrm{\angle }Q\sin \mathrm{\angle }S\sin \mathrm{\angle }P$$$$=2r^2\sin \frac{\pi -\alpha }{2}\sin \frac{\pi -\beta }{2}\sin \frac{\pi -\gamma }{2}$$$$=2r^2\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\cos \frac{\gamma }{2}$$$$\frac{S_{SPQ}}{S_{ABC}}=\frac{2r^2\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\cos \frac{\gamma }{2}}{R(\sin \alpha +\sin \beta +\sin \gamma )r}$$$$=\frac{2r\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\cos \frac{\gamma }{2}}{R\times 4\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\cos \frac{\gamma }{2}}$$$$=\frac{r}{2R}✓$$

Commented by mnjuly1970 last updated on 01/Dec/22

$$veryniceproof...thanksalot$$

Answered by mnjuly1970 last updated on 01/Dec/22

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