Question 251610
consider the triangle BCD which is a right angled triangle

now using Pythagoras theorem we get 
     {{{BC^2=CD^2 + BD^2}}}------------1

now consider the triangle ACD which is also a right angled triangle

again using Pythagoras th. 
       {{{AC^2=AD^2+CD^2}}}------------2

but AB=AC (isosceles) 

so substituting it in 2 we get
       {{{AB^2=AD^2+CD^2}}}

but AB=AD+DB

so,  {{{(AD+BD)^2=AD^2+CD^2}}}

i.e., {{{BD^2+AD^2+2*AD*BD=AD^2+CD^2}}}

on simplifying we get {{{CD^2=BD^2+2*AD*BD}}}

substituting the value of CD^2 in equation 1

{{{BC^2=BD^2+2*AD*BD + BD^2}}}

taking out BD common in the right hand side of the equation 

{{{BC^2=2*BD*(AD+BD)}}}
but AD+BD=AB

therefore {{{BC^2=2AB*BD}}}