Question no. 3 Share with your friends Share 1 Varun.Rawat answered this GIVEN :Two circle with centres O and O' intersect at 2 points A and B. AB is the common chord of 2 circles and OO' is the line segment joining the centres of the two circles. Let OO'intersect AB at P.TO PROVE :OO' is the ⊥ bisector of ABCONSTRUCTION :Join OA, OB, O'A and O'BPROOF :In ∆OAO' and OBO'OO' = OO' CommonOA = OB Radii of same circleO'A = O'B Radii of same circle∆OAO' ≅ ∆OBO' SSS⇒∠AOO' = ∠BOO' CPCT⇒∠AOP = ∠BOPIn ∆AOP and ∆BOP,OP = OP Common∠AOP = ∠BOP Proved aboveOA = OB Radii of same circle∆AOP ≅ ∆BOP SAS⇒AP = BP CPCT⇒∠APO = ∠BPO CPCTBut ∠APO + ∠BPO =180° Linear pair⇒∠APO + ∠APO = 180°⇒∠APO = 90° = ∠BPOThus, AP = BP and ∠APO = ∠BPO = 90°°Hence, OO' is the ⊥ bisector of AB. 5 View Full Answer
