prove that the line joining the mid point of a chord to the center of the circle passes through the mid point of the corresponding minor arc
Let AB be a chord of a circle having centre at O. Let C be the mid-point of chord AB.
We know that the line the centre of a circle to the chord is always perpendicular to the chord.
∴ OC ⊥ AB
So, OC be the perpendicular bisector of the chord AB and on extending it will cut the circle at P and Q respectively.
Now, in triangle OAC and OBC, we have
OA = OB [radii of same circle]
OC = OC [common]
and ∠OCA = ∠OCB [Each being a right angle]
∴ Δ OCA Δ OCB [SAS congruency]
⇒∠AOC = ∠BOC [c.p.c.t]
⇒ minor arc AQ = minor arc BQ