AB and AC are two chord of a circle whose radius is r. if p and q are the distance of chord AB and CD from the Central respectively and AB=2BC than prove that 4q*=p*+3r*

AB and AC are two chord of a circle whose radius is r. if p and q are

1.	AB and AC are two chord of a circle whose radius is r. if p and q are the distance of chord AB and CD from the Central respectively and AB=2BC than prove that 4q=p+3r*
Answer» Given, AB and AC are two chords of a circle of radius rAgain, AB = 2AC and the distances of AB and AC from the center are p and q.Let AC = x, then AB = 2xWe know that the perpendicular drawn from the centre to the chord bisect the chord.So, CL = x/2 and AM = xNow, in ΔOLC,r2= q2+ (x/2)2=> r2= q2+ x2/4Again, in ΔOAM, 4q2= 4(r2- x2/4) (using 1) 4r2= 4r2- x2=> 4q2= 4r2- (r2- p2) (using 2)=> 4q2= 4r2- r2+ p2=> 4q2= 3r2+ p2