Question

Prove that the perpendicular bisector of a chord of a circle is the bisector of the corresponding arc of the circle.

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Let AB be the chord of a circle that has centre at O.

Let the mid-point of chord AB = C

Since, the line to the centre of a circle to the chord is always perpendicular to the chord.  Therefore, OC ⊥ AB

Also, OC is the perpendicular bisector of the chord AB and will cut the circle at P and Q.

In Δ OAC and Δ OBC

OA = OB  [radii of the same circle]

OC = OC  [common]

∠OCA = ∠OCB  [right angles]  

Therefore, Δ OCA ~ Δ OCB  [SAS congruency]

= ∠AOC = ∠BOC  

= Minor arc AQ = Minor arc BQ