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^2
this question is related to lesson circles

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 .    <=== ! ! !