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.