Asked by monicasyal | 31st Jan, 2010, 10:12: PM
Given: Two circles with centre A and B intersect each other at point P and Q.
PQ is the common chord to the circles.
To Prove: AB is the perpendicular bisector of a common chord
Proof: We know that the line from the centre is perpendicular bisector of the chord.
The line BL passing from centre B, BLP = 90o .................(1)
PL = PQ. ......................(2)
The line AL passing from centre A, ALP = 90o .............(3)
Now, BLP + ALP = 180o (From (1) and (3))
i.e AB is staright line with L as a point on it.
From (1), (2) and (3)
The line AB is a perpendicular bisector of the common chord PQ
Answered by | 1st Feb, 2010, 10:44: AM
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