# 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

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 . <=== ! ! !