Pls answer it dont provide link

Hi, 
Assuming that point D is anywhere on line BC 


Let BD = x such that CD = 2-x 
now given ,
$$\begin{gathered} {\mathrm{BD}}^2= \mathrm{BC}\hspace{0.17em}\times \mathrm{DC} \\ {\mathrm{x}}^2= 2\times \left(2-\mathrm{x}\right) \\ {\mathrm{x}}^2= 4-2\mathrm{x} \\ {\mathrm{x}}^2+2\mathrm{x}-4 = 0 \\ \mathrm{Using} \mathrm{quadratic} \mathrm{formula}, \\ \mathrm{x} = \frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ = \frac{-2\pm \sqrt{4+16}}{2}= -1\pm \sqrt{5} \\ \mathrm{now} \mathrm{length} \mathrm{cannot} \mathrm{be} \mathrm{negative} \mathrm{so} \\ \mathrm{x} = \sqrt{5}-1 \\ \mathrm{so} \mathrm{length} \mathrm{of} \mathrm{DC} = 2-\mathrm{x} = 2-\left(\sqrt{5}-1 \right)=\left( 3-\sqrt{5}\right) \mathrm{cm} \end{gathered}$$