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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