Given that \(\vec a\)and \(\vec b\)are two vectors such that |\(\vec a\)+ \(\vec b\)| = |\(\vec a\)|.

Now, |\(\vec a\) + \(\vec b\)|² = (\(\vec a\) + \(\vec b\)) ∙ (\(\vec a\) + \(\vec b\)) = \(\vec a\) ∙ \(\vec a\) + \(\vec a\) ∙ \(\vec b\) + \(\vec b\) ∙ \(\vec a\) + \(\vec b\) ∙ \(\vec b\) 

= |\(\vec a\)|² + 2\(\vec a\) ∙ \(\vec b\) + \(\vec b\) ∙ \(\vec b\). 

( \(\because\) \(\vec a\)∙ \(\vec a\) = |\(\vec a\)|² and \(\vec a\) ∙ \(\vec b\)= \(\vec b\)∙ \(\vec a\)) 

⇒ |\(\vec a\) + \(\vec b\)|² = |\(\vec a\) + \(\vec b\)|² + 2\(\vec a\) ∙ \(\vec b\) + \(\vec b\) ∙ \(\vec b\)( \(\because\) |\(\vec a\) + \(\vec b\)| = |\(\vec a\)| (Given)) 

⇒ 2\(\vec a\) ∙ \(\vec b\)+ \(\vec b\) ∙ \(\vec b\) = 0. ... (1)

Now, (2\(\vec a\)+ \(\vec b\)) ∙ \(\vec b\) = 2\(\vec a\) ∙ \(\vec b\)+ \(\vec b\)∙ \(\vec b\)= 0 (By equation (1) ) 

We know that two vectors \(\vec a\) and \(\vec b\) are perpendicular only if  \(\vec a\) ∙ \(\vec b\) = 0. 

Hence, that vector (2\(\vec a\) + \(\vec b\)) is perpendicular to the vector \(\vec b\).