Mathematics
Prove that the perpendicular bisector of a chord of a circle always passes through the centre.
Answer
AB is a chord of a circle with centre O.
Let CD be the perpendicular bisector of AB.
∠ACD = 90° [CD ⊥ AB]
Line joining the center of a circle to the midpoint of a chord is perpendicular to the chord.
∠ACO = 90° [Since C is the midpoint of AB, OC ⊥ AB]
∴ ∠ACD = ∠ACO which is wrong
∴ CD must pass through O.
Hence, the perpendicular bisector of a chord of a circle always passes through the centre.
Related Questions
Show that equal chords of a circle subtend equal angles at the centre of the circle.
In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at P. If L and M are mid-points of AB and CD respectively, prove that OLPM is a square.
AB and CD are two parallel chords of a circle and a line l is the perpendicular bisector of AB. Show that l is the perpendicular bisector of CD also.
Prove that diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.