Question

please solve the both Question given in attachment

Answer 1

Let two circles O and O' intersect at two points A and B so that AB is the common chord of two circles and OO' is the line segment joining the centres Let OO' intersect AB at M Now Draw line segments OA, OB , O'A and O'B In ΔOAO' and OBO' , we have OA = OB (radii of same circle) O'A = O'B (radii of same circle) O'O = OO' (common side) ⇒ ΔOAO' ≅ ΔOBO' (SSS congruency) ⇒ ∠AOO' = ∠BOO' ⇒ ∠AOM = ∠BOM ......(i) Now in ΔAOM and ΔBOM we have OA = OB (radii of same circle) ∠AOM = ∠BOM (from (i))OM = OM (common side) ⇒ ΔAOM ≅ ΔBOM (SAS congruncy) ⇒ AM = BM and ∠AMO = ∠BMO But ∠AMO + ∠BMO = 180° ⇒ 2∠AMO = 180° ⇒ ∠AMO = 90°Thus, AM = BM and ∠AMO = ∠BMO = 90° Hence OO' is the perpendicular bisector of AB.

Answer 2

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