Chord of a circle:
The line segment joining any two points on the circumference of the circle is known as chord of the circle. Diameter is the longest chord of circle which passes through center of the circle. The figure below depicts a circle and its chord.
In the given circle with ‘O’ as center, AB represents the diameter of the circle (longest chord) , ‘OE’ denotes radius of the circle and CD represents a chord of the circle.
Let us consider the chord CD of the circle and two points P and Q anywhere on the circumference of the circle except the chord as shown. If the end points of the chord CD are joined to the point P, then the angle ∠CPD is known as the angle subtended by the chord CD at point P. The angle ∠ CQD is the angle subtended by chord CD at Q. The angle ∠ COD is the angle subtended by chord CD at the center O.
If we try to establish a relationship between different chords and the angle subtended by them on the center, we see that the longer chord subtends a greater angle at the center. Similarly two chords of equal length subtend equal angle at the center. Let us try to prove this statement.
Theorem 1: Chords which are equal in length subtend equal angles at the center of the circle.
Proof: From fig. 3, In ∆AOB and ∆POQ
S.No. |
Statement |
Reason |
1. |
AB=PQ |
Chords of equal length (Given) |
2. |
OA = OB = OP = OQ |
Radius of the same circle |
3. |
\(\bigtriangleup AOB = \bigtriangleup POQ\) |
SSS axiom of Congruence |
4. |
\(\angle AOB = \angle POQ\) |
From Statement 3 |
Note: CPCT stands for congruent parts of congruent triangles.
Converse of theorem 1 also holds true, which states that if two angles subtended by two chords at the center are equal then the chords are of equal length. From fig. 3, if∠AOB =∠POQ, then AB=PQ. Let us try to prove this statement.
Theorem 2:If the angles subtended by the chords of a circle are equal in measure then the length of the chords is equal.
Proof:From fig. 3, In ∆AOB and ∆POQ
S.No. |
Statement |
Reason |
1. |
\(\angle AOB = \angle POQ\) |
Equal angle subtended at centre O (Given) |
2. |
OA = OB = OP = OQ |
Radii of the same circle |
3. |
\(\bigtriangleup AOB \cong \bigtriangleup POQ\) |
SAS axiom of Congruence |
4. |
AB = PQ |
From Statement 3 (CPCT) |
Theorem 3: Equal chords of a circle are equidistant from the center of the circle.
Given: Chords AB and CD are equal in length.
Construction: Join A and C with centre O and drop perpendiculars from O to the chords AB and CD.
Construction: Join A and C with centre O and drop perpendiculars from O to the chords AB and CD.
S.No |
Statement |
Reason |
1 |
\(AP = \frac{AB}{2}\) |
Perpendicular from centre bisect the chord |
In \(\bigtriangleup OAP \;\; and \;\; \bigtriangleup OCQ\) |
||
2 |
\(\angle 1 = \angle 2 = 90^{\circ}\) |
\(OP \perp AB \;\; and \;\; OQ \perp CD\) |
3 |
\(OA = OC\) |
Radii of the same circle |
4 |
\(OP = OQ\) |
Given |
5 |
\(\bigtriangleup OPB \cong \bigtriangleup OQD\) |
R.H.S. Axiom of Congruency |
6 |
AP = CQ |
Corresponding parts of congruent triangle |
7 |
AB = CD |
From statement (1) and (6) |
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