$ABCD$ is a square where each side is a uniform wire of resistance $1\,\Omega$ . $A$ point $E$ lies on $CD$ such that if a uniform wire of resistance $1\,\Omega$ is connected across $AE$ and constant potential difference is applied across $A$ and $C$ then $B$ and $E$ are equipotential.
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Let resistance $D E=x \Omega$ and that between $C E=(1-x) \Omega .$ The points $\mathrm{B}$ and $\mathrm{E}$ will be equipotential, when effective resistance acros $\mathrm{AE}=$ effective, resistance across $CE$
i.e., $\frac{(1+x) \times 1}{(1+x)+1}=(1-x)$ or $(1+x)=(1-x) \times(2+x)$
or $1+x=2+x-2 x 2 x-x^{2}$ or $x^{2}+2 x-1=0$
On solving it, we get, $x=\sqrt{2}-1$
Now, $1-x=2-(\sqrt{2}-1)=2-\sqrt{2}=(\sqrt{2}-1)$
$\therefore \frac{C E}{E D}=\frac{1-x}{x}=\frac{\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2}-1)}=\sqrt{2}$
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