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Theorem 2 - The Perpendicular from the Centre of a Circle to a Chord Bisects the Chord

Theorem 2 : The perpendicular from the centre of a circle to a chord bisects the chord.
Given : PQ is the chord of circle C(O, r) and OL
PQ.
To Prove : PL = QL
Construction : Join OP and OQ.

 

Proof :
In
ΔOPL and ΔOQL
OL=OL (common)
OP=OQ=r (radii of same circle)

OLP=OLQ (each 90°)
...   ΔOPL ≅ ΔOQL
...   PL = QL

 

Example :

If a line l intersects two concentric circles at the points P, Q, R and S as shown in the figure, prove that PQ = RS.

Solution :

Let O be the centre of the two concentric circles. Draw OT l. QR is the chord of the smaller circle and PS is the chord of the bigger circle.
So, QT = RT      (i)
and PT = ST     (ii)
...  PT - QT = ST - RT       (from ii - i )
PQ = RS







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