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Geometry Rectangle Formed by two Chords
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Chord of a Circle:Example
Theorem
The perpendicular from the center of a circle to a chord bisects the chord.
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Example: Prove that the perpendicular bisector of a chord of a circle always passes through the centre.
Solution: Let PQ be a chord of a circle having its centre at O.
Let AB be the perpendicular bisector of chord PQ. If possible, suppose AB does not pass through O. Then, PAO is a part of PAB. Since AB is the perpendicular bisector of chord PQ
PAB = 90 ...(i)
Since A is mid-point of PQ and O is the centre of the circle.
PAO = 90 ...(ii) 
From (i) and (ii), you get PAO = PAB
This is a contradiction.
Hence, AB must pass through O.
 
   
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