Question

prove that the perpendicular bisector of a chord of a circle always passes through the centre

Answer 1

Well, this theorem can be proved very easily using logic. But, i am not sure, whether your subject teacher will accept it or not. Besides all these, we have to just activate our logic. We will see step by step.

GIVEN : A circle has the chord AB which is bisected by OP, i.e., OP is the perpendicular bisector of the chord AB and as a result, AP = BP.

TO PROVE : O is the centre of the circle through which the perpendicular bisector of ghe chord AB will pass.

CONSTRUCTION : Join O,A and O,B.

PROOF : In ∆APO & ∆BPO,

AP = BP (•.• OP bisects AB)

<OPA = <OPB = 90° (•.• OP ⊥ AB)

OP is the common side

Hence, ∆ APO is congruent to the other ∆ BPO.

Therefore, OA = OB (C.P.C.T)

Now, centre of a circle is the only fixed point from where other points on the circumference of that circle are equidistant. Since, A and B are equidistant from O, O is clearly the centre of the circle. Also, AP = BP (i.e., OP is the perpendicular bisector of AB) is possible if and only if OP passes through "O", which is the centre of the circle.

Hence, the theorem is proved.

Answer 2

Solution: Let AB be a chord

of a circle having its centre

at O .

Let PQ be the perpendicular

bisector of chord AB.

If possible , suppose PQ

doesn't pass through O.

Then, <APO is a part of <APQ.

Since, PQ is the perpendicular

bisector of chord AB.

<APQ = 90° -------( 1 )

Since, P is midpoint of AB and

O is the centre of the circle .

<APO = 90° -------( 2 )

From (1) and (2) , we get

<APO = <APQ

This is a contradiction.

Hence, PQ must pass through 'O'

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