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

  • 48
What are you looking for?