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Question Number 207985 by efronzo1 last updated on 02/Jun/24

Answered by mr W last updated on 02/Jun/24

$$sayAC=s=AB=CB$$$${AP}^2=s\times AM\Rightarrow AP=\sqrt{s\times AM}$$$${CP}^2=s\times CN\Rightarrow CP=\sqrt{s\times CN}$$$$AP+CP=\sqrt{s}(\sqrt{AM}+\sqrt{CN})=s$$$$\Rightarrow s={(\sqrt{AM}+\sqrt{CN})}^2={\left(2\sqrt{2}\right)}^2=8✓$$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{I}\mathrm{saw}\mathrm{the}\mathrm{theorem}\mathrm{here}.$$$$\mathrm{Thanks}\mathrm{sir}.$$

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