Question

All the theorems of circles class 9 -Maths 9th

Answer 1

The Most important theorems (10) in Circle are Below ... Note: The derivations are also given below. Theorem: Equal chords of a circle subtend equal angles at the centre. Theorem: This is the converse of the previous theorem. It implies that if two chords subtend equal angles at the center, they are equal. Theorem: A perpendicular dropped from the center of the circle to a chord bisects it. It means that both the halves of the chords are equal in length. . Theorem: The line that is drawn through the center of the circle to the midpoint of the chords is perpendicular to it. In other words, any line from the center that bisects a chord is perpendicular to the chord.  Theorem: If there are three non-collinear points, then there is just one circle that can pass through them.  Theorem: Equal chords of a circle are equidistant from the center of a circle.  Theorem: This is the converse of the previous theorem. It states that chords equidistant from the center of a circle are equal in length. Theorem: The angle subtended by an arc at the center of a circle is double that of the angle that the arc subtends at any other given point on the circle.  Theorem: Angles formed in the same segment of a circle are always equal in measure. Theorem: If the line segment joining any two points subtends equal angles at two other points that are on the same side, they are concyclic. This means that they all lie in the same circle. Important Information or knowledge required : Segment and Sector of the Circle A segment of the circle is the region between either of its arcs and a chord. It could be a major or minor segment. Sector of the circle is the area covered by an arc and two radii joining the centre of the circle. It could be the major or minor sector. Angle Subtended by a Chord at a Point If in a circle AB is the chord and is making ∠ACB at any point of the circle then this is the angle subtended by the chord AB at a point C.  Likewise, ∠AOB is the angle subtended by chord AB at point O i.e. at the centre and ∠ADB is also the angle subtended by AB at point D on the circle. _________________________________________________ THEOREMS  Theorem 1   Statement: Equal chords of a circle subtend equal angles at the centre.  Given: AB and CD are chords of a circle with centre O, such that AB = CD.  To prove: ∠AOB = ∠COD Proof:  In △AOB and △COD,  AO = CO (radii of the same circle) BO = DO (radii of the same circle) AB = CD (given)  ∴ △AOB ≅ △COD (SSS)  Hence, ∠AOB = ∠COD (c.p.c.t.) Theorem 2 Statement: If the angles subtended by the chords of a circle at the centre are equal, then the chords are  equal. Given: Two chords PQ and RS of a circle C(O, r), such that ∠POQ = ∠ROS.  To prove: PQ = RS Proof: In △POQ and △ROS,  OP = OQ = OR = OS = r (radii of the same circle) and ∠POQ = ∠ROS (given) ∴ △POQ ≅ △ROS (SAS) ∴ PQ = RS. (c.p.c.t.)  Theorem 3 Statement: The perpendicular from the centre of a circle to a chord bisects the chord. Given: AB is the chord of a circle with centre O and OD ⊥ AB.  To prove: AD = DB  Construction: Join OA and OB. Proof:  In △ODA and △ODB,  OA = OB (radii of the same circle) OD = OD (common)  ∠ODA = ∠ODB (each is a rt. angle) △ODA ≅ △ODB (R.H.S.) AD = DB (c.p.c.t.)  Theorem 4 Statement: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Given: A chord PQ of a circle C(O, r) and L is the mid-point of PQ.  To prove: OL ⊥ PQ.  Construction: Joint OP and OQ. Proof: In △OLP and △OLQ, OP = OQ (radii of the same circle) PL = QL (given) OL = OL (common) ∴ △OLP ≅ △OLQ (SSS) Also, ∠OLP + ∠OLQ = 180° (linear pair) ∴ ∠OLP = ∠OLQ = 90° Hence, OL ⊥PQ Theorem 5 Statement: There is one and only one circle passing through three given non-collinear points. Theorem 6 Statement: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres). Given: AB and CD are two equal chords of a circle.  OM and ON are perpendiculars from the centre to the chords AB and CD. To prove: OM = ON. Construction: Join OA and OC. Proof: In △AOM and △CON, OA = OC (radii of the same circle) MA = CN (Since, OM and ON are perpendicular to  the chords and it bisects the chord and AM = MB, CN = ND) ∠OMA = ∠ONC = 90° ∴ △AOM ≅ △CON (R.H.S.) ∴ OM = ON (c.p.c.t.) Equal chords of a circle are equidistant from the centre. Theorem 7 Statement: Chords equidistant from the centre of a circle are equal in length. Given: OM and ON are perpendiculars from the centre to the chords AB and CD and OM = ON. To prove: Chord AB = Chord CD. Construction: Join OA and OC. Proof:   OM ⊥ AB ⇒ 1/2 AB = AM  ON ⊥ CD ⇒ 1/2 CD = CN Consider △AOM and △CON,  OA = OC (radii of the same circle) OM = ON (given)  ∠OMA = ∠ONC = 90° (given) △AOM ≅ △CON (RHS congruency) AM = CN ⇒ 1/2 AB = 1/2 CD ⇒ AB = CD The two chords are equal if they are equidistant from the centre.  Theorem 8 Statement: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.  Given: O is the centre of the circle. To prove: ∠BOC = 2∠BAC Construction: Join O to A. Proof: In △AOB, OA = OB (radii of the same circle)  ⇒ ∠1 = ∠2 Similarly in △AOC, ∠3 = ∠4 Now, by exterior angle property, ∠5 = ∠1 + ∠2 ∠6 = ∠3 + ∠4 ⇒ ∠5 + ∠6 = ∠1 + ∠2 + ∠3 + ∠4 ⇒ ∠5 + ∠6 = 2∠2 + 2∠3  = 2(∠2 + ∠3) ⇒ ∠BOC = 2∠BAC Theorem 9 Statement: Angles in the same segment of a circle are equal. Given: Two angles ∠ACB and ∠ADB are in the same segment of a circle C(O, r). To prove: ∠ACB = ∠ADB Construction: Join OA and OB. Proof:  In fig. (i), we know that, angle subtended by an arc of a circle at the centre is double the angle subtended by the  arc in the alternate segment.  Hence, ∠AOB = 2∠ACB  ∠AOB = 2∠ADB So, ∠ACB = ∠ADB In fig. (ii), we have,  Reflex ∠AOB = 2∠ACB  and Reflex ∠AOB = 2∠ADB 2∠ACB = 2∠ADB ∴ ∠ACB = ∠ADB Theorem 10 Statement: If a line segment joining two points subtends equal angles at two other points lying on the  same side of the line containing the line segment, the four points lie on a circle (i.e. they are  concyclic).  Theorem 11 Statement: The sum of either pair of opposite angles of a cyclic quadrilateral is 180°. Given: Let ABCD be a cyclic quadrilateral To prove: ∠A + ∠C = 180° and ∠B + ∠D = 180° Construction: Join OB and OD. Proof:  ∠BOD = 2∠BAD  ∠BAD = 1/2 ∠BOD Similarly, ∠BCD = 1/2 reflex ∠BOD  ∴ ∠BAD + ∠BCD = 1/2 ∠BOD + 1/2 reflex ∠BOD = 1/2 (∠BOD + reflex ∠BOD) = 1/2 ×360° ∴ ∠A + ∠C = 180° Similarly, ∠B + ∠D = 180° Theorem 12 Statement: If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.