Results of Equal Chords

**Theorem 1: Equal chords are equidistant from the center.**

Given:

Construction: Join and

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

Since

Similarly,

Given

Now consider and

(proved above)

(radius)

Therefore (By RHS criterion)

. Hence proved that equal chords are equidistant from the center.

**Theorem 2 (Converse of Theorem 1): Chords of a circle which are equidistant from the center are equal.**

Given: Two chords and in circle

Also where and

To Prove:

Construction: Join and

Proof: (given)

We know that the perpendicular from the center of a circle to a chord, bisects the chord.

Similarly

Consider and

(radius)

(given)

Therefore (By S.A.S criterion)

Hence proved that chords of a circle which are equidistant from the center are equal.

**Theorem 3: Equal chords of concurrent circles are equidistant from the corresponding centers.**

Given: Two concurrent circles and and

Construction: Join and

To Prove: where and

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

Similarly,

Now (given)

Consider and

(congruent circles with radius )

(proved above)

Therefore (By RHS criterion)

Therefore i.e. and are equidistant from and respectively.

**Theorem 4 (Converse of Theorem 3): Chords of congruent circles which are equidistant from their corresponding centers are equal.**

Given: Two circles and . and are chords such that . and

To Prove:

Construction: Join and

Proof: Consider and

(given)

(equal radius)

Therefore (By RHS criterion)

We know that the perpendicular from the center of a circle to a chord, bisects the chord.

Hence proved that chords of congruent circles which are equidistant from their corresponding centers are equal.

**Theorem 5: Equal chords of a circle subtends equal angles at the center.**

Given: and are two equal chords in circle

To Prove:

Proof: Consider and

(radius)

(radius)

(given)

Therefore

.

Hence proved that equal chords of a circle subtend equal angles at the center.

**Theorem 6: (converse of Theorem 5) If angles subtended by two chords of a circle at the center are equal, then the chords are equal.**

Given: and are chords of a circle such that

To Prove:

Proof: Consider and

(radius)

(radius)

(given)

Therefore (By S.A.S criterion)

Therefore (corresponding sides off congruent triangles are equal)

**Theorem 7: Equal chords of congruent triangles subtend equal angles at the center.**

Given: Two concurrent circles and . Chords

To Prove:

Proof: Consider and

(radius)

(radius)

(given)

Therefore (By S.S.S criterion)

Hence (corresponding angles of congruent triangles are equal).

**Theorem 8 (converse of Theorem 7): If the angles subtended by the two chords of congruent circles at the corresponding centers of circles are equal, the chords are equal.**

Given: Two congruent circles and . AB and CD are chords such that

To Prove:

Proof: Consider and

(radius)

(radius)

(given)

Therefore (By S.A.S criterion)

**Theorem 9: Of any two chords of a circle, the chord which is larger is nearer to the circle.**

Given: Two chords and of a circle with center such that

Construction: Join and

To Prove: or

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

And

In right triangles and , we have

And

Since (radius)

Given

(Adding to both sides)

Therefore

Hence proved that of any two chords of a circle, the chord which is larger is nearer to the circle.

**Theorem 10 (converse of Theorem 9): Of any two chords of a circle, the chord which is nearer is larger.**

Given: Two chords and of a circle such that , where and are perpendicular from on and respectively.

To Prove:

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

Therefore

In right triangles and we have

And

… … … … … i)

And … … … … … ii)

Now

(Adding on both sides)

Since (radius of the circle)

. Hence proved that of any two chords of a circle, the chord which is nearer is larger.