Question

solve questions 2 and 3... ​

Answer 1

Draw OP⊥AB and O'Q⊥CD. Consider ΔOPM and ΔO'QM OM = O'M [Given]∠OMP = ∠O'QM [Vertically opposite angles] ∠OPM = ∠O'QM = 90° [Construction] ∠POM = ∠QO'M [Angle sum property of a triangle] So, ΔOPM ≅ ΔO'QM OP = O'Q [CPCT] ⇒ AB = CD [Equal chords of same circle or equal circles are equidistance from the centres of the respective circles.] Hence proved.