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Question Number 183633 by a.lgnaoui last updated on 27/Dec/22

$$Montrerque$$$${\left(\frac{a}{b}\right)}^2=\frac{c}{d}+1$$

Commented by a.lgnaoui last updated on 27/Dec/22

Answered by mr W last updated on 28/Dec/22

Commented by mr W last updated on 28/Dec/22

$$R=\frac{c+d}{2}$$$${(R-r)}^2-{(R-r-d)}^2=b^2$$$$\Rightarrow d(2R-2r-d)=b^2$$$$\Rightarrow d(c-2r)=b^2$$$$\Rightarrow 2r=c-\frac{b^2}{d}$$$${\left(\sqrt{a^2+r^2}\right)}^2-{(R+R-r-d)}^2=b^2$$$$a^2+r^2-{(c-r)}^2=b^2$$$$a^2-c^2+2cr=b^2$$$$a^2-c^2+c(c-\frac{b^2}{d})=b^2$$$$a^2=(1+\frac{c}{d})b^2$$$$\Rightarrow {\left(\frac{a}{b}\right)}^2=\frac{c}{d}+1✓$$

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