PLEASE HELP ME WITH IT.In the figure given below , AB and AC are two chords of a circle of center O and radius r. If AB= 2 AC and the perpendiculars drawn from the center on these chords are of lengths 'a' and 'b' respectively. PROVE THAT 4b^2= a^2+3r^2This question is related to lesson CIRCLES

• AL2006:
  OK.  I did it.  Now let's see if I can go through it without
  getting too complicated.
  
  I think the key to the whole thing is this fact:
  
       A radius drawn perpendicular to a chord bisects the chord.
  
  That tells us several things:
  
  -- OM bisects AB. 
     'M' is the midpoint of AB.
     AM is half of AB.
  
  -- ON bisects AC.
      'N' is the midpoint of AC.
     AN is half of AC.
  
  --  Since AC is half of AB,
       AN is half of AM.
       a = b/2 
  
  Now look at the right triangle inside the rectangle.
  'r' is the hypotenuse, so
  
                                              a² + b² = r²
  
  But  a = b/2, so             (b/2)² + b² = r²
  
  (b/2)² = b²/4                   b²/4   + b² = r²
  
  Multiply each side by 4:     b² + 4b² = 4r²
                                         -  -  -  -  -  -  -  -  -  -  -
                                              0  + 5b² = 4r²  
  Repeat the
  original equation:                a² +  b² =  r²
  
  Subtract the last
  two equations:                  -a² + 4b² = 3r² 
  
  Add  a²  to each side:              4b²  =  a² + 3r² .    <=== ! ! !