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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. There is one and only one circle passing through three given non-collinear points.

**Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.**

Given: A circle with centre O. AC is a chord and OB âŠ¥ AC.

To prove: AB = BC.

Construction: Join OA and OC.

Proof: In triangles OBA and OBC,

âˆ OBA = âˆ OBC = 90^{o} (Since OB âŠ¥ AC)

OA = OC (Radii of the same circle)

OB = OB (Common side)

Î”OBA â‰… Î”OBC (By RHS congruence rule)

â‡’ AB = BC (Corresponding sides of congruent triangles)

Thus, OB bisects the 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: A circle with centre O. AC is a chord and AB = BC.

To prove: OB âŠ¥ AC.

Construction: Join OA and OC.

Proof: In triangles OBA and OBC,

AB = BC (Given)

OA = OC (Radii of the same circle)

OB = OB (Common side)

Î”OBA â‰… Î”OBC (SSS congruence rule)

â‡’ âˆ OBA = âˆ OBC (Corresponding angles of congruent triangles)

But, âˆ OBA + âˆ OBC = âˆ ABC = 180Â° [Linear pair]

âˆ OBC + âˆ OBC = 180Â°^{ }[Since âˆ OBA = âˆ OBC]

2 x âˆ OBC = 180Â°

âˆ OBC = $\frac{\text{180\xc2\xb0}}{\text{2}}$ = 90^{o}

âˆ OBC = âˆ OBA = 90Â°

âˆ´ OB âŠ¥ AC

Hence, the theorem is proved.

Let AB and PQ be any two chords of a circle with the centre O. âˆ AOB and âˆ POQ are called the angles subtended by the chord at the centre of the circle. As the chord moves away from the centre, its length and the angle subtended by it at the centre decreases. On the other hand, if a chord moves closer to the centre, its length and the angle subtended by it at the centre increases.

Given: A circle with centre O. AB and PQ are chords of the circle. AB = PQ

To prove: âˆ AOB = âˆ POQ

Proof: In triangles AOB and POQ,

AB = PQ (Given)

OA = OP (Radii of same circle)

OB = OQ (Radii of same circle)

Î”AOB â‰… Î”POQ (SSS congruence rule)

â‡’ âˆ AOB = âˆ POQ (Corresponding angles)

Hence, the theorem is proved.

**Theorem: Chords that subtend equal angles at the centre of a circle are equal in length.**

Given: âˆ AOB = âˆ POQ

To prove: AB = PQ

Proof: In triangles AOB and POQ,

âˆ AOB = âˆ POQ (Given)

OA = OP (Radii of same circle)

OB = OQ (Radii of same circle)

Î”AOB â‰… Î”POQ (SAS congruence rule)

â‡’ AB = PQ (Corresponding sides)

Hence, the theorem is proved.

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. There is one and only one circle passing through three given non-collinear points.

**Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.**

Given: A circle with centre O. AC is a chord and OB âŠ¥ AC.

To prove: AB = BC.

Construction: Join OA and OC.

Proof: In triangles OBA and OBC,

âˆ OBA = âˆ OBC = 90^{o} (Since OB âŠ¥ AC)

OA = OC (Radii of the same circle)

OB = OB (Common side)

Î”OBA â‰… Î”OBC (By RHS congruence rule)

â‡’ AB = BC (Corresponding sides of congruent triangles)

Thus, OB bisects the 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: A circle with centre O. AC is a chord and AB = BC.

To prove: OB âŠ¥ AC.

Construction: Join OA and OC.

Proof: In triangles OBA and OBC,

AB = BC (Given)

OA = OC (Radii of the same circle)

OB = OB (Common side)

Î”OBA â‰… Î”OBC (SSS congruence rule)

â‡’ âˆ OBA = âˆ OBC (Corresponding angles of congruent triangles)

But, âˆ OBA + âˆ OBC = âˆ ABC = 180Â° [Linear pair]

âˆ OBC + âˆ OBC = 180Â°^{ }[Since âˆ OBA = âˆ OBC]

2 x âˆ OBC = 180Â°

âˆ OBC = $\frac{\text{180\xc2\xb0}}{\text{2}}$ = 90^{o}

âˆ OBC = âˆ OBA = 90Â°

âˆ´ OB âŠ¥ AC

Hence, the theorem is proved.

Let AB and PQ be any two chords of a circle with the centre O. âˆ AOB and âˆ POQ are called the angles subtended by the chord at the centre of the circle. As the chord moves away from the centre, its length and the angle subtended by it at the centre decreases. On the other hand, if a chord moves closer to the centre, its length and the angle subtended by it at the centre increases.

Given: A circle with centre O. AB and PQ are chords of the circle. AB = PQ

To prove: âˆ AOB = âˆ POQ

Proof: In triangles AOB and POQ,

AB = PQ (Given)

OA = OP (Radii of same circle)

OB = OQ (Radii of same circle)

Î”AOB â‰… Î”POQ (SSS congruence rule)

â‡’ âˆ AOB = âˆ POQ (Corresponding angles)

Hence, the theorem is proved.

**Theorem: Chords that subtend equal angles at the centre of a circle are equal in length.**

Given: âˆ AOB = âˆ POQ

To prove: AB = PQ

Proof: In triangles AOB and POQ,

âˆ AOB = âˆ POQ (Given)

OA = OP (Radii of same circle)

OB = OQ (Radii of same circle)

Î”AOB â‰… Î”POQ (SAS congruence rule)

â‡’ AB = PQ (Corresponding sides)

Hence, the theorem is proved.