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^2 This question is related to lesson CIRCLES

(1) Answers

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 .    <=== ! ! !  

Add answer