Question

proof of the theorem the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord

Answer 1

Given, a circle with center O, in which line OM bisect the chord AB, i.e., AM = MB.

To prove: OM ⊥ AB

Construction: Join OA and OB.

Proof:

In triangle OAM and OBM, we have

OA = OB  (radii of same circle)

OM = OM

Also, AM = BM (given)

therefore triangle OAM and OBM are congruent.

⇒∠OMA = ∠OMB  [c.p.c.t]

Now, ∠OMA + ∠OMB = 180°  [linear pair]

⇒ 2∠OMA = 180°

⇒ ∠OMA = 90°

Hence, OM ⊥ AB

Answer 2

Statement : The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Given : A chord PQ of a circle C ( O, r ) and L is the mid - point of PQ.

To prove : OL ⊥ PQ

Construction : Joint OP and OQ.

PROOF :

In ∆OLP and ∆OLQ,

OP = OQ . (radii of same circle)
PL = QL . (given)
OL = OL . (common)

ΔOLP ≅ ΔOLQ . (SSS)

Also,

∠OLP + ∠OLQ = 180° (linear pair)
∠OLP = ∠OLQ = 90°

Hence, OL ⊥ PQ.