Geometry Rectangle Formed by two Chords

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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