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Question Number 206484 by sonukgindia last updated on 16/Apr/24

Answered by mr W last updated on 16/Apr/24

Commented by mr W last updated on 16/Apr/24

$$R=\frac{AD}{2}=\frac{m+n+x}{2}$$$$CF=\sqrt{n^2+R^2-nR}$$$$\frac{AE}{FD}=\frac{AC}{CF}$$$$\Rightarrow a=\frac{(m+x)R}{\sqrt{n^2+R^2-nR}}$$$$similarly$$$$\Rightarrow b=\frac{(n+x)R}{\sqrt{m^2+R^2-mR}}$$$$a^2+b^2={\left(2R\right)}^2$$$$\frac{{(m+x)}^2{R}^2}{n^2+R^2-nR}+\frac{{(n+x)}^2{R}^2}{m^2+R^2-mR}=4R^2$$$$\frac{{(m+x)}^2}{n^2+R^2-nR}+\frac{{(n+x)}^2}{m^2+R^2-mR}=4$$$$\frac{{(m+x)}^2}{4n^2+{(m+n+x)}^2-2n(m+n+x)}+\frac{{(n+x)}^2}{4m^2+{(m+n+x)}^2-2m(m+n+x)}=1$$$$\frac{{(m+x)}^2}{{(m+x)}^2+3n^2}+\frac{{(n+x)}^2}{{(n+x)}^2+3m^2}=1$$$$\frac{1}{1+3{\left(\frac{n}{m+x}\right)}^2}+\frac{1}{1+3{\left(\frac{m}{n+x}\right)}^2}=1$$$$1=9{\left(\frac{n}{m+x}\right)}^2{\left(\frac{m}{n+x}\right)}^2$$$$1=\frac{3mn}{(m+x)(n+x)}$$$$(m+x)(n+x)=3mn$$$$\Rightarrow x^2+(m+n)x-2mn=0✓$$

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