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Question Number 199688 by cortano12 last updated on 07/Nov/23

Commented by cortano12 last updated on 07/Nov/23

$$\mathrm{prove}\mathrm{that}\mathrm{formula}$$

Answered by deleteduser1 last updated on 07/Nov/23

$$Letdandebethediagonalsofthecyclic$$$$quadrilateral;$$$${Ptolemy}^{\prime }stheorem\Rightarrow de=2br+ac...\left(i\right)$$$$Also,d=\sqrt{4r^2-a^2};e=\sqrt{4r^2-c^2}$$$$\Rightarrow de=\sqrt{16r^4-4r^2{c}^2-4a^2{r}^2+a^2{c}^2}=2br+ac$$$$\Rightarrow 16r^4-4r^2{c}^2-4a^2{r}^2+a^2{c}^2=4b^2{r}^2+a^2{c}^2+4abcr$$$$\Rightarrow 4r^2(4r^2-c^2-a^2)=4r^2(b^2+\frac{abc}{r})$$$$\Rightarrow a^2+b^2+c^2+\frac{abc}{r}=4r^2$$

Answered by mr W last updated on 07/Nov/23

Commented by mr W last updated on 07/Nov/23

$$\beta =\pi -\alpha \Rightarrow \cos \beta =-\cos \alpha =-\frac{c}{2r}$$$$a^2+b^2+2ab\times \frac{c}{2r}={AC}^2={\left(2r\right)}^2-c^2$$$$\Rightarrow a^2+b^2+c^2+\frac{abc}{r}=4r^2$$

Answered by som(math1967) last updated on 07/Nov/23

$$cos\mathrm{\angle }BAC=\frac{a}{2r}$$$$cos(180-\mathrm{\angle }BAC)=\frac{b^2+c^2-{BD}^2}{2bc}$$$$\frac{a}{2r}=\frac{{BD}^2-b^2-c^2}{2bc}$$$$\Rightarrow \frac{a}{r}=\frac{4r^2-a^2-b^2-c^2}{bc}$$$$\Rightarrow \frac{abc}{r}+a^2+b^2+c^2=4r^2$$

Commented by som(math1967) last updated on 07/Nov/23

Answered by MM42 last updated on 07/Nov/23

$${AB}^2=a^2+b^2-2abcos\alpha$$$${AB}^2=4r^2-c^2$$$$\Rightarrow a^2+b^2+c^2-2abcos\alpha =4r^2$$$$\Rightarrow a^2+b^2+c^2-2abcos(\pi -\beta )=4r^2$$$$\Rightarrow a^2+b^2+c^2+2abcos\beta =4r^2$$$$\Rightarrow a^2+b^2+c^2+\frac{abc}{r}=4r^2✓$$$$$$

Commented by MM42 last updated on 07/Nov/23

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