The perpendicular bisector of two chord of a circle ibsercet of its centre
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Answer:
Answer:
Given: PQ and RS are two chords of a circle C(O, r) and their mid points L and M respectively. Suppose perpendicular bisectors O'L and O'M of PQ and RS respectively meet at O'.
To prove : (O' coincides with O)
Construction : Join OL and OM
Proof : Since L is the mid-point of chord PQ, therefore
\sf\qquad{OL\perp PQ}OL⊥PQ
OL is the perpendicular bisector of PQ.
OL as well as O'L are perpendicular bisectors of PQ.
O'L lies along OL............(I)
Similarly, M is mid-point of RS
OM \perp⊥ RS.
OM is the perpendicular bisector of RS.
OM as well as O'M are perpendicular bisectors of RS.
O'M lies along OM............(II)
From (i) and (ii), we find that O'L and O'M lie along OL and OM respectively. This means that the point of intersection of O'L and O'M coincides with the point of intersection of OL and OM i.e. O' coincides with O.