# In a triangle ABC,Dis a point on BC such that 9BD=BC,then relation between AB and AD

Answer :  ABD is a triangle

$\\ AD^2=AB^2+BD^2 -2(AB)(BD)CosB \\\\\Rightarrow AD^2=AB^2+(BC/9)^2-2(AB)(BC/9)CosB$

Now

$\\ \Rightarrow (AD/AB)^2= 1+1/9(BC/AB)^2-2/9(BC/AB)CosB$

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