Tow persons A and B move …

Let two men start from the point C with velocity v each at the same time.
Also, {tex}\angle BCA = {45^o}{/tex} 
Since, A and B are moving with same velocity v, so they will cover same distance in same time.
Therefore, {tex}\triangle ABC{/tex} is an isosceles triangle with AC = BC.
Now, draw {tex}CD \bot AB{/tex} 
Let at any instant t, the distance between them is AB

Let AC = BC = x and AB = y
In {tex}\triangle ACD{/tex} and {tex}\triangle DCB{/tex},
{tex}\angle CAD = \angle CBD{/tex} {tex}\left[ {\because AC = BC} \right]{/tex} 
{tex}\angle CDA = \angle CDB = {90^o}{/tex} 
{tex}\therefore \angle ACD = \angle DCB{/tex} 
or {tex}\angle ACD = \frac{1}{2} \times \angle ACB{/tex} {tex} \Rightarrow \angle ACD = \frac{1}{2} \times {45^o}{/tex} 
{tex} \Rightarrow \angle ACD = \frac{\pi }{8}{/tex} 
{tex}\therefore \Rightarrow \sin \frac{\pi }{8} = \frac{{AD}}{{AC}}{/tex}
{tex}\therefore \Rightarrow \sin \frac{\pi }{8} = \frac{{y/2}}{x}\,\left[ {\because AD = \frac{y}{2}} \right]{/tex} 
{tex}\Rightarrow \frac{y}{2} = x\sin \frac{\pi }{8}{/tex} 
{tex} \Rightarrow y = 2x\sin \frac{\pi }{8}{/tex} 
Now, differentiating both sides w.r.t t, we get
{tex}\frac{{dy}}{{dx}} = 2.\sin \frac{\pi }{8}.\frac{{dx}}{{dt}}{/tex} 
{tex}= 2.\sin \frac{\pi }{8}.v\,\left[ {\because v = \frac{{dx}}{{dt}}} \right]{/tex} 
{tex}= 2v.\frac{{\sqrt {2 - \sqrt 2 } }}{2}\,\,\left[ {\because \sin \frac{\pi }{8} = \frac{{\sqrt {2 - \sqrt 2 } }}{2}} \right]{/tex} 
{tex} = \sqrt {2 - \sqrt 2 } {/tex} v unit /s
which is the rate at which A and B are being separated.