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CBSE - IX
Chords of a Circle
Lesson Demo
You know that the perpendicular from a point to a line segment is the shortest distance between them. A line that joins two points on the circumference of a circle is called a chord. A chord passing through the centre of a circle is called the diameter. The longest chord of a circle is the diameter.
Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.
Given: OB ⊥ AC.
To prove: AB = BC.
Construction: Join OA and OC.
Proof: In triangles OBA and OBC,
∠OBA = ∠OBC = 90o (Since OB ⊥ AC.)
OA = OC (Radii of same circle)
OB = OB (Common side)
ΔOBA 
ΔOBC (By RHS congruence rule)
AB = BC
Thus, OB bisects chord AC.
Hence, the theorem is proved.
Theorem: The line drawn from the centre of a circle to bisect a chord is perpendicular to the chord.
Given: AB = BC
To prove: OB ⊥ AC.
Construction: Join OA and OC.
Proof: In triangles OBA and OBC,
AB = BC (Given)
OA = OC (Radii of same circle)
OB = OB (Common side)
Δ OBA Δ OBC (SSS congruence rule)
∠OBA = ∠OBC
∠OBA + ∠OBC = ∠ABC = 180o
Since ∠OBA = ∠OBC,
2 x ∠OBC = 180o
∠OBC = 
= 90o
∠OBC = 90o = ∠OBA
∴OB ⊥ AC
