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Question-207985




Question Number 207985 by efronzo1 last updated on 02/Jun/24
Answered by mr W last updated on 02/Jun/24
say AC=s=AB=CB  AP^2 =s×AM ⇒AP=(√(s×AM))  CP^2 =s×CN ⇒CP=(√(s×CN))  AP+CP=(√s)((√(AM))+(√(CN)))=s  ⇒s=((√(AM))+(√(CN)))^2 =(2(√2))^2 =8 ✓
$${say}\:{AC}={s}={AB}={CB} \\ $$$${AP}^{\mathrm{2}} ={s}×{AM}\:\Rightarrow{AP}=\sqrt{{s}×{AM}} \\ $$$${CP}^{\mathrm{2}} ={s}×{CN}\:\Rightarrow{CP}=\sqrt{{s}×{CN}} \\ $$$${AP}+{CP}=\sqrt{{s}}\left(\sqrt{{AM}}+\sqrt{{CN}}\right)={s} \\ $$$$\Rightarrow{s}=\left(\sqrt{{AM}}+\sqrt{{CN}}\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{8}\:\checkmark \\ $$
Commented by Tawa11 last updated on 21/Jun/24
I saw the theorem here.  Thanks sir.
$$\mathrm{I}\:\mathrm{saw}\:\mathrm{the}\:\mathrm{theorem}\:\mathrm{here}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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