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      Class 9 Maths

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      • Class 09
      • Class 9 Maths
      CoursesClass 09MathsClass 9 Maths
      • 1.Number System
        14
        • Lecture1.1
          Numbers and its Classification, Rational Number and its Representation on the Number Line 50 min
        • Lecture1.2
          Decimal Representation of Rational Number and Conversion of Decimal Number into Rational Number 01 hour
        • Lecture1.3
          Representation of Decimal Number on Number Line, Finding Irrational Numbers Between Given Two Numbers 46 min
        • Lecture1.4
          Representing Irrational Number on the Number Line 41 min
        • Lecture1.5
          Rationalisation 54 min
        • Lecture1.6
          Introduction to Exponents, Laws of Exponents, BODMAS Rule 41 min
        • Lecture1.7
          Integral Exponents of Real Numbers 57 min
        • Lecture1.8
          Integral Exponents of Real Numbers and Solving Equation for x 01 hour
        • Lecture1.9
          Rational Exponents of Real Numbers 52 min
        • Lecture1.10
          Examples Based on Rational Exponents of Real Numbers 30 min
        • Lecture1.11
          Chapter Notes – Number System
        • Lecture1.12
          NCERT Solutions – Number System Exercise 1.1 – 1.6
        • Lecture1.13
          R D Sharma Solutions Number System
        • Lecture1.14
          Revision Notes – Number System
      • 2.Polynomials
        10
        • Lecture2.1
          Introduction to Polynomials and Some Basics Questions 58 min
        • Lecture2.2
          Value of Polynomial, Zeroes of Polynomial and Relationship between Zeroes and Coefficient of Linear Equation 01 hour
        • Lecture2.3
          Division of Polynomials and Division Algorithm and Remainder Theorem 51 min
        • Lecture2.4
          Factorising by Spliting Method 01 hour
        • Lecture2.5
          Factorisation By Different Methods and Factor Theorem 27 min
        • Lecture2.6
          Factorisation of Polynomial by Using Algebric Identities 01 hour
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.5
        • Lecture2.9
          R S Aggarwal Solutions Polynomials
        • Lecture2.10
          Revision Notes Polynomials
      • 3.Coordinate Geometry
        8
        • Lecture3.1
          Representation of Position on Number Line, Finding the location by Using the Axis 01 hour
        • Lecture3.2
          Quadrant, Coordinates of Point on Axis, Distance from Axis 01 hour
        • Lecture3.3
          Finding the Area of Triangle and Rectangle and Mirror Image of Point On Axis 57 min
        • Lecture3.4
          Chapter Notes – Coordinate Geometry
        • Lecture3.5
          NCERT Solutions – Coordinate Geometry Exercise 3.1 – 3.3
        • Lecture3.6
          R D Sharma Solutions Coordinate Geometry
        • Lecture3.7
          R S Aggarwal Solutions Coordinate Geometry
        • Lecture3.8
          Revision Notes Coordinate Geometry
      • 4.Linear Equations
        8
        • Lecture4.1
          Introduction of Linear Equation in Two Variables and Graphical Representation 01 hour
        • Lecture4.2
          Example to find the value of Constant and Graph of Linear Equation in Two Variables 57 min
        • Lecture4.3
          Equations of lines Parallel to the x-axis and y-axis 46 min
        • Lecture4.4
          Chapter Notes – Linear Equations
        • Lecture4.5
          NCERT Solutions – Linear Equations Exercise 4.1 – 4.4
        • Lecture4.6
          R D Sharma Solutions Linear Equations
        • Lecture4.7
          R S Aggarwal Solutions Linear Equations
        • Lecture4.8
          Revision Notes Linear Equations
      • 5.Euclid's Geometry
        7
        • Lecture5.1
          Euclid’s Definitions and Euclid’s Axioms 32 min
        • Lecture5.2
          Eulcid’s Postulates and Playfair’s Axiom 37 min
        • Lecture5.3
          Chapter Notes – Euclid’s Geometry
        • Lecture5.4
          NCERT Solutions – Euclid’s Geometry Exercise 5.1 5.2
        • Lecture5.5
          R D Sharma Solutions Euclid’s Geometry
        • Lecture5.6
          R S Aggarwal Solutions Euclid’s Geometry
        • Lecture5.7
          Revision Notes Euclid’s Geometry
      • 6.Lines and Angles
        10
        • Lecture6.1
          Introduction to Lines and Angles, Types of Angle and Axioms 01 hour
        • Lecture6.2
          Vertically opposite Angles and Angle bisector 46 min
        • Lecture6.3
          Angle made by a transversal with two Lines and Parallel Lines 01 hour
        • Lecture6.4
          Lines Parallel to same Line; Exterior angle of a triangle; Angle Sum property of a Triangle 52 min
        • Lecture6.5
          Angle Sum Property of Triangle Including Interior and Exterior Angles 01 hour
        • Lecture6.6
          Chapter Notes – Lines and Angles
        • Lecture6.7
          NCERT Solutions – Lines and Angles Exercise 6.1 – 6.3
        • Lecture6.8
          R D Sharma Solutions Lines and Angles
        • Lecture6.9
          R S Aggarwal Solutions Lines and Angles
        • Lecture6.10
          Revision Notes Lines and Angles
      • 7.Triangles
        11
        • Lecture7.1
          Introduction, Types of Triangle, Altitude & Median of Triangle, Congruences, SAS criterion of Congruence 42 min
        • Lecture7.2
          Questions Based on SAS Criterion 01 hour
        • Lecture7.3
          ASA Criterion of Congruence and Its based Questions 42 min
        • Lecture7.4
          Questions Based on AAS and SSS Criterion 29 min
        • Lecture7.5
          RHS Criterion of Congruence and Its based Questions, Questions Based on Multiple Criterion 01 hour
        • Lecture7.6
          Inequalities of Triangle 42 min
        • Lecture7.7
          Inequalities of Triangle cont., Distance Between a line and The Point, Miscellaneous Questions 48 min
        • Lecture7.8
          Chapter Notes – Triangles
        • Lecture7.9
          NCERT Solutions – Triangles Exercise 7.1 – 7.5
        • Lecture7.10
          R D Sharma Solutions Triangles
        • Lecture7.11
          Revision Notes Triangles
      • 8.Quadrilaterals
        13
        • Lecture8.1
          Congruence of Triangles- SAS Congruence Criteria, ASA Congruence Criteria, SSS Congruence Criteria, RHS Congruence Criteria; Types of Quadrilateral- Parallelogram, Rectangle, Rhombus, Square, Trapezium, Kite 01 hour
        • Lecture8.2
          Theorem 1; Properties of parallelogram- Theorem 2, Theorem 3; 01 hour
        • Lecture8.3
          Theorem 4; Theorem 5; 24 min
        • Lecture8.4
          Conditions for a Quadrilateral to be a parallelogram- Theorem 6 01 hour
        • Lecture8.5
          Theorem 8 01 hour
        • Lecture8.6
          Theorem 9 51 min
        • Lecture8.7
          Theorem 9(Mid-point Theorem); 01 hour
        • Lecture8.8
          Theorem 10 27 min
        • Lecture8.9
          Chapter Notes – Quadrilaterals
        • Lecture8.10
          NCERT Solutions – Quadrilaterals Exercise 8.1 8.2
        • Lecture8.11
          R D Sharma Solutions Quadrilaterals
        • Lecture8.12
          R S Aggarwal Solutions Quadrilaterals
        • Lecture8.13
          Revision Notes Quadrilaterals
      • 9.Area of Parallelogram
        11
        • Lecture9.1
          Parallelograms on the Same Base and between the Same Parallels; A Diagonal of a parallelogram divides it into two triangles of Equal Area 57 min
        • Lecture9.2
          Questions based on parallelogram and Triangle are on same base & parallel lines and Area of Triangle 01 hour
        • Lecture9.3
          Triangles on the same Base and between the same Parallels 56 min
        • Lecture9.4
          Miscellaneous Questions Based on Triangles on the same Base and between the same Parallels 54 min
        • Lecture9.5
          Sums Based on Median of a Triangle 52 min
        • Lecture9.6
          Two Triangles on the same Base and equal Area lie b/w the same parallel 01 hour
        • Lecture9.7
          Summary of All Concepts and Miscellaneous Questions 56 min
        • Lecture9.8
          Chapter Notes – Area of Parallelogram
        • Lecture9.9
          NCERT Solutions – Area of Parallelogram Exercise 9.1 – 9.4
        • Lecture9.10
          R D Sharma Solutions Area of Parallelogram
        • Lecture9.11
          Revision Notes Area of Parallelogram
      • 10.Constructions
        7
        • Lecture10.1
          Construction of Perpendicular Bisector of a Line Segment, Different Angles and Angle bisector 42 min
        • Lecture10.2
          Construction of Different types of Triangle 43 min
        • Lecture10.3
          Chapter Notes – Constructions
        • Lecture10.4
          NCERT Solutions – Constructions Exercise
        • Lecture10.5
          R D Sharma Solutions Constructions
        • Lecture10.6
          R S Aggarwal Solutions Constructions
        • Lecture10.7
          Revision Notes Constructions
      • 11.Circles
        11
        • Lecture11.1
          Introduction, Important Terms related to circle, Theorem-1, 2, 3 and 4 01 hour
        • Lecture11.2
          Problems Solving 48 min
        • Lecture11.3
          Problems Based on Parallel Chords, Common Chord of Circles 01 hour
        • Lecture11.4
          Problems Based on Concentric Circles, Triangle in Circle 41 min
        • Lecture11.5
          Equal Chords and Their Distance, Theorem-5, 6, 7, 8 01 hour
        • Lecture11.6
          Circle Through Points, Theorem-9, Arc of a Circle, Theorem-10 01 hour
        • Lecture11.7
          Chapter Notes – Circles
        • Lecture11.8
          NCERT Solutions – Circles Exercise
        • Lecture11.9
          R D Sharma Solutions Circles
        • Lecture11.10
          R S Aggarwal Solutions Circles
        • Lecture11.11
          Revision Notes Circles
      • 12.Heron's Formula
        8
        • Lecture12.1
          Introduction of Heron’s Formula; Area of Different Triangles 01 hour
        • Lecture12.2
          Area of Triangles Using Heron’s Formula (Type-1 level-1 and 2) 53 min
        • Lecture12.3
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-1) 32 min
        • Lecture12.4
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-2) 38 min
        • Lecture12.5
          Chapter Notes – Heron’s Formula
        • Lecture12.6
          NCERT Solutions – Heron’s Formula Exercise 12.1 12.2
        • Lecture12.7
          R D Sharma Solutions Heron’s Formula
        • Lecture12.8
          Revision Notes Heron’s Formula
      • 13.Surface Area and Volume
        16
        • Lecture13.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere 40 min
        • Lecture13.2
          Type-1: Surface Area and Volume of Cube and Cuboid 27 min
        • Lecture13.3
          Level-2, Question based on Cube and Cuboid, Level-1 Questions based on Cylinder 28 min
        • Lecture13.4
          Level-2 Questions based on Cylinder, Type-3 Surface Area and Volume of Right Circular Cone (Level-1) 25 min
        • Lecture13.5
          Level-2 Questions based on Right Circular Cone, Level-1& 2 Questions Based on Sphere & Hemisphere 27 min
        • Lecture13.6
          Type-5 Short Answer Types Questions, 12 min
        • Lecture13.7
          Chapter Notes – Surface Area and Volume of a Cuboid and Cube
        • Lecture13.8
          Chapter Notes – Surface Area and Volume of A Right Circular Cylinder
        • Lecture13.9
          Chapter Notes – Surface Area ad Volume of A Right Circular Cone
        • Lecture13.10
          Chapter Notes – Surface Area ad Volume of A Sphere
        • Lecture13.11
          NCERT Solutions – Surface Area and Volume Exercise 13.1 – 13.9
        • Lecture13.12
          R D Sharma Solutions Surface Areas And Volumes of A cuboid and A cube
        • Lecture13.13
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cylinder
        • Lecture13.14
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cone
        • Lecture13.15
          R D Sharma Solutions Surface Areas And Volumes of A Sphere
        • Lecture13.16
          Revision Notes Surface Areas And Volumes
      • 14.Statistics
        15
        • Lecture14.1
          Data; Presentation of Data; Understanding Basic terms; 01 hour
        • Lecture14.2
          Questions; 52 min
        • Lecture14.3
          Bar Graph, Histogram, Frequency Polygon; 01 hour
        • Lecture14.4
          Histogram for Grouped Frequency Distribution (Class interval in Inclusive form), Frequency Polygon, Measures of Central Tendency, 01 hour
        • Lecture14.5
          Examples Based on Mean of Ungrouped Data 56 min
        • Lecture14.6
          Direct Method of Finding Mean of Ungrouped Data, Median of Cont. Freq. Distribution (Grouped Data) 01 hour
        • Lecture14.7
          Mean; Mean for an ungrouped freq. Dist. Table 25 min
        • Lecture14.8
          Chapter Notes – Statistics – Tabular Representation of Statistical Data
        • Lecture14.9
          Chapter Notes – Statistics – Graphical Representation of Statistical Data
        • Lecture14.10
          Chapter Notes – Statistics – Measures of Central Tendency
        • Lecture14.11
          NCERT Solutions – Statistics Exercise 14.1 – 14.4
        • Lecture14.12
          R D Sharma Solutions Tabular Representation of Statistical Data
        • Lecture14.13
          R D Sharma Solutions Graphical Representation of Statistical Data
        • Lecture14.14
          R S Aggarwal Solutions Statistics
        • Lecture14.15
          Revision Notes Statistics
      • 15.Probability
        8
        • Lecture15.1
          Activities; Understanding term Probability; Experiments; 44 min
        • Lecture15.2
          Event; Probability; Empirical Probability- Tossing a coin; 31 min
        • Lecture15.3
          Rolling a dice; Miscellaneous Example; 27 min
        • Lecture15.4
          Chapter Notes – Probability
        • Lecture15.5
          NCERT Solutions – Probability Exercise 15.1
        • Lecture15.6
          R D Sharma Solutions Probability
        • Lecture15.7
          R S Aggarwal Solutions Probability
        • Lecture15.8
          Revision Notes Probability

        Chapter Notes – Circles

        (1) A circle is the collection of those points in a plane that are at a given constant distance from a fixed-point in the plane. The fixed point is called the centre and the given constant distance is called the radius of the circle.
        A Circle with centre O and radius r usually denoted by C(O,r). Thus, in set theoretic notations, we write C(O,r)={X:OX=r}

        (2) A point P lies inside or on or outside the circle C(O,r) according as OP<r or OP=r or OP>r.

        (3) The collection of all points lying inside and on the circle C(O,r) is called a circular disc with centre O and radius r.
        The set of all points lying inside and on the circle is called a Circular Disc. It is also known as the circular region.

        (4) Circles having the same center and different radii are said to be concentric circles.
        When two or more circles have the same center but have different radii, they are called as concentric circles, that is, circles with common center.

        (5) A continuous piece of a circle is called an arc of the circle.
        For Example: Consider circle C (O, r). Let P1,P2,P3,P4,P5,P6 be point on the circle. Then, the pieces P1,P2,P3,P4,P5,P6,P1,P2 etc. are all arcs of the circle C(O,r).

         

        (6) Prove that If two arcs of circle are congruent,  then corresponding chords are equal.

        Given: Arc PQ of a Circle C(O,r) and arc RS of another circle C(O′,r) such that PQ≅RS
        To Prove: PQ=RS
        Construction: Draw Line segment OP, OQ, O′R and O′S.Proof:
        Case-I When arc(PQ) and arc(RS) are minor Arcs
        In triangle OPQ and O′RS, We have
        OP=OQ=O′R=O′S=r                   [Equal radii of two circles]
        ∠POQ=∠RO′S                 arc(PQ)≅arc(RS)⇒m(arc(PQ))≅m(arc(RS))⇒∠POQ=∠RO′S
        So by SAS Criterion of congruence, we have
        ΔPOQ≅ΔRO′S
        ⇒PQ=RS
        Case-II When arc(PQ) and arc(RS) are major arcs.
        If arc(PQ), arc(RS) are major arcs, then arc(QP) and arc(SR) are Minor arcs.
        So arc(PQ)≅arc(RS)
        ⇒ arc(QP)≅arc(SR)
        ⇒QP=SR
        ⇒PQ=RS
        Hence, PQ≅RS ⇒PQ=RS

        (7) Prove that If two chords of a circle are equal, then their corresponding arcs are congruent.

        Given: Equal chords, PQ of a circle C(O,r)  and RS of congruent circle C(O′,r)
        To Prove: arc(PQ)≅arc(RS), where both arc(PQ) and arc(RS) are minor, major or semi-circular arcs.
        Construction: If PQ,RS are not diameters, draw line segments OP, OQ, O′R and O′S.Proof:
        Case I: when arc(PQ) and arc(RS) are diameters
        In this case, PQ and RS are semi circle of equal radii, hence they are congruent.
        Case II: When arc(PQ) and arc(RS) are Minor arcs.
        In triangles POQ and RO′S, we have
        PQ=RS
        OP=O′R=r and OQ=O′S=r
        So by SSS-criterion of congruence, we have
        ΔPOQ≅ΔRO′S
        ⇒            ∠POQ=∠RO′S
        ⇒           m(arc(PQ))=m(arc(RS))
        ⇒         arc(PQ)≅arc(RS)
        Case III: When arc(PQ) and arc(RS) are major arcs
        In this case, arc(QP) And arc(SR)will be minor arcs.
        PQ=RS
        ⇒            QP=SR
        ⇒           m(arc(QP))=m(arc(SR))
        ⇒           360∘−m(arc(PQ))−360∘−m(arc(RS))
        ⇒           m(arc(PQ))−m(arc(RS))
        ⇒           arc(PQ)≅arc(RS)
        Hence, in all the three cases, we have arc(PQ)≅arc(RS)

         

        (8) Prove that The perpendicular from the centre of a circle to a chord bisects the chord.

        Given: A Chord PQ of a circle C(O,r) and perpendicular OL to the chord PQ.
        To Prove: LP=LQ
        Construction: Join OP and OQProof: In Triangles PLO and QLO, we have
        OP=OQ=r                               [Radii of the same circle]
        OL=OL                                       [Common]
        And,      ∠OLP=∠OLQ     [Each equal to 90∘]
        So, by RHS-criterion of congruence, we have
        ΔPLO≅ΔQLO
        ⇒            PL=LQ

        (9) Prove that The line segment joining the centre of a circle to the mid-point of a chord is perpendicular to the chord.

        Given: A Chord PQ OF a circle C(O,r) with mid-point M.
        To Prove: OM⊥PQ
        Construction: Join OP and OQProof: In triangles OPM and OQM, we have
        OP=OQ                                     [Radii of the same circle]
        PM=MQ                                   [M is mid-point of PQ]
        OM=OM
        So, by SSC- criterion of congruence, we have
        ΔOPM≅ΔOQM
        ⇒            ∠OMP=∠OMQ
        But , ∠OMP+∠OMQ=180∘                                          [Linear pair]
        ⇒            ∠OMP+∠OMP=180∘  [∠OMP=∠OMQ]
        ⇒            2∠OMP=180∘
        ⇒            ∠OMP=90∘

        (10) Prove that There is one and only circle passing through three given points.

        Given: Three non-collinear points P,Q and R.
        To Prove: There is one and only one circle passing through P,Q and R.
        Construction: Join PQ and QR. Draw perpendicular bisectors AL and BM of PQ and RQ respectively. Since P,Q and R. are not collinear. Therefore, the perpendicular bisectors AL and BM are not parallel.
        Let AL And BM intersect at O. Join OP,OQ and OR.

        Proof: Since O lies on the perpendicular bisector of PQ.
        Therefore,
        OP=OQ
        Again, O Lies on the perpendicular bisector of QR.
        Therefore,
        OQ=OR
        Thus,     OP=OQ=OR=r    (say)
        Taking O as the centre draw a circle of radius s. Clearly, C(O,s) passes through P, Q and R. This proves that there is a circle passing the points P,Q and R.
        We shall now prove that this is the only circle passing through P,Q and R.
        If possible, let there be another circle with centre O′ and radius r, passing through the points P,Q and R. Then, O′ will lie on the perpendicular bisectors AL of PQ and BM of QR.
        Since two lines cannot intersect at more than one point, so O′ must coincide with O. Since OP=r, O′P=s and O and O′ coincide, we must have
        r=s
        ⇒           C(O,r)=C(O′,s)
        Hence, there is one and only one circle passing through three non-collinear points P,Q and R.

         

        (11) Prove that If two circles intersect in two points, then the through the centre is perpendicular to the common chord.

        Given: Two circles C(O,r) and C(O′,s) intersecting at points A and B.
        To Prove: OO′ is perpendicular bisector of AB.
        Construction: Draw line segments
        OA,OB,O′A and O′BProof: In triangles OAO′ and OBO′,  we have
        OA=OB=r
        O′A=O′B=s
        And,                OO′=OO′
        So, by SSS-criterion of congruence, we have
        ΔOAO≅ΔOBO′
        ⇒       ∠AOO′=∠BOO′
        ⇒       ∠AOM=∠BOM         [∠AOO′=∠AOM and ∠BOM=∠BOO′]
        Let M be the point of intersection of AB and OO′
        In triangles AOM and BOM, we have
        OA=OB=r
        ⇒       ∠AOO′=∠BOO′
        ⇒       ∠AOM=∠BOM        [∠AOO′=∠AOM and ∠BOM=∠BOO′]
        Let M be the point of intersection of AB and OO′
        In triangles AOM and BOM, we have
        OA=OB=r
        ∠AOM=∠BOM
        And                 OM=OM
        So, by SAS-criterion of congruence, we have
        ΔAOM≅ΔBOM
        ⇒       AM=BM and ∠AMO=∠BOM
        But,                             ∠AOM+∠BMO=180∘
        2∠AOM=180∘
        ⇒       ∠AOM=90∘
        Thus,                           AM=BM and ∠AOM=∠BMO=90∘
        Hence, OO′ is the perpendicular bisector of AB.

        (12) Prove that Equal chords of a circle subtend equal angle at the centre.

        Given: Two Chord AB and CD of circle C(O,r) such that AB=CDand OL⊥AB and OM⊥CD
        To Prove: Chord AB and CD are equidistant from the centre O i.e OL=OM.
        Construction: Join OA and OC.Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord.
        Therefore,
        OL⊥AB ⇒ AL=12AB……….(i)
        And,                OM⊥CD ⇒ CM=12CD……….(ii)
        But,                 AB=CD
        ⇒       12AB=12CD
        ⇒       AL=CM                                [Using (i) and (ii) ]……….(iii)
        Now, in right triangles OAL and OCM, we have
        OA=OC                                           [Equal to radius of the circle]
        AL=CM                                            [From equation (iii)]
        And,                ∠ALO=∠CMO         [Each equal to 90∘]
        So by RHS criterion of convergence, we have
        ΔOAL≅ΔOCM
        ⇒       OL=OM
        Hence, equal chord of a circle are equidistant from the centre.

        (13) Prove that Chords of a circle which are equidistant from the centre are equal.

        Given: Two Chords AB and CD of a circle C(O,r) which are equidistant from its centre i.e. OL=OM, where OL⊥AB and OM⊥CD.
        To Prove: Chords are Equal i.e. AB=CD
        Construction: Join OA and OCProof: Since the perpendicular from the centre of a circle to a chord bisects the chord.
        Therefore,
        OL⊥AB
        ⇒       AL=BL
        ⇒       AL=12AB
        And,                            OM⊥CD
        ⇒       CM=DM
        ⇒       CM=12CD
        In triangles OAL and OCM, we have
        OA=OC                        [Each equal to radius of the given Circle]
        ∠OLA=∠OMC         [Each equal to 90∘]
        And,                OL=OM                                            [Given]
        So, by RHS, criterion of convergence, we have
        ΔOAL≅ΔOCM
        ⇒       AL=CM
        ⇒       12AL=12AB
        ⇒       AB=CD
        Hence, the chords of a circle which are equidistant from the centre are equal.

         

        (14) Prove that Equal chords of a circle subtend equal angle at the centre.

        Given: A circle C(O,r) and its two equal chords AB and CD.
        To Prove: ∠AOB=∠CODProof:In triangles AOB and COD, we have
        AB=CD                                                [Given]
        OA=OC                                     [Each equal to r]
        OB=OD                                                [Each equal to r]
        So, by SSC-criterion of Congruence, we have
        ΔAOB≅ΔCOD
        ⇒       ∠AOB=∠COD

        (15) Prove that If the angles subtended by two chords of a circle at the centre are equal, the chords are equal.

        Given: Two Chord AB and CD of a circle C(O,r) such that ∠AOB=∠COD
        To Prove: AB=CDProof: In triangles AOB and COD, we have
        OA=OC                                     [Each equal to r]
        ∠AOB=∠COD         [Given]
        OB=OD                                     [Each equal to r]
        So, by SAS-criterion of congruence, we have
        ΔAOB≅ΔCOD
        ⇒       AB=CD

        (16) Prove that Of any two chords of a circle, the larger chord is nearer to the centre.

        Given: Two Chord AB and CD of a circle with Centre O such that AB>CD
        To Prove: Chord AB is nearer to the centre of the circle i.e. OL<OM, where OL and OM are perpendiculars from O to AB and CD respectively
        Construction: Join OA and OC.Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord. Therefore,
        OL⊥AB ⇒ AL=12AB
        And,                OM⊥CD ⇒ CM=12CD
        In right triangles OAL and OCM, we have
        OA2=OL2+AL2
        And,                OC2=OM2+CM2
        ⇒       OL2+AL2=OM2+CM2.. (i)  [OA=OC⇒OA2=OC2]
        Now,                           AB>CD
        ⇒       12AB>12CD
        ⇒       AL>CM
        ⇒       AL2>CM2
        ⇒       OL2+AL2>OL2+CM2         [Adding OL2 on both sides]
        ⇒       OM2+CM2>OL2+CM2       [using equation (i)]
        ⇒       OM2>OL2
        ⇒       OM>OL
        ⇒       OL<OM
        Hence, AB is nearer to the centre than CD.

        (17) Prove that Of any two chords of a circle, the chord nearer to the centre is larger.

        Given: Two Chord AB and CD of a circle C(O,r) such that OL<OM, where OL and OM are perpendiculars From O on AB and CD respectively.
        To Prove: AB>CD
        Construction: Join OA and OC.Proof: Since the perpendicular From the Centre of a circle to a chord bisects the chord.
        AL=12AB and CM=12CD
        In right triangles OAL and OCM, we have
        OA2=OL2+AL2 and, OC2=OM2+CM2
        ⇒       AL2=OA2−OL2……. (i)
        And,                            CM2=OC2−OM2…….(ii)
        Now,                           OL<OM
        ⇒       OL2<OM2
        ⇒       −OL2>−OM2
        ⇒       OA2−OL2>OA2−OM2         [adding OA2 on both sides]
        ⇒       OA2−OL2>OC2−OM2          [OA2=OC2]
        ⇒       AL2>CM2
        ⇒       AL>CM
        ⇒       2AL>2CM
        ⇒       AB>CD

        (18) Prove that The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

        Given: An arc PQ of a circle C(O,r) and a point R on the remaining part of the circle i.e. arc QP.
        To Prove: ∠POQ=2∠PRQ
        Construction: join RO and produce it to a point M outside the circle.

        Proof: We shall consider the following three different cases:
        Case I: when arc(PQ) is a minor arc.
        We know that an exterior angle of a triangle is equal to the sum of the interior oppsite angles.
        In ΔPOQ, ∠POM is the exterior angle.
        ∠POM=∠OPR+∠ORP
        ⇒ ∠POM=∠ORP+∠ORP     [OP=OR=r, ∠OPR=∠ORP]⇒ ∠POM=2∠ORP…………………(i)
        In ΔQOR, ∠QOM is the exterior angle.
        ∠QOM=∠OQR+∠ORQ
        ⇒ ∠QOM=∠OQP+∠ORQ   [OQ=OR=r, ∠ORQ=∠OQR]
        ⇒ ∠QOM=2∠ORQ…………….(ii)
        Adding equation (i) and (ii), we get
        ∠POM+∠QOM=2∠ORP+2∠ORP
        ⇒ ∠POM+∠QOM=2(∠ORP+∠ORP)
        ⇒  ∠POM=2∠PRQ
        Case II: when arc(PQ) is a semi-circle
        We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles.
        In ΔPOQ, we have
        ∠POM=∠OPR+∠ORP
        ⇒ ∠POM=∠ORP+∠ORP     [OP=OR=r, ∠OPR=∠ORP]
        ⇒ ∠POM=2∠ORP…………………(iii)
        In ΔQOR, We have
        ∠QOM=∠ORQ+∠OQR
        ⇒ ∠QOM=∠ORQ+∠ORQ   [OQ=OR=r, ∠ORQ=∠OQR]
        ⇒ ∠QOM=2∠ORQ…………….(iv)
        Adding equations (iii) and (iv), we get
        ∠POM+∠QOM=2(∠ORP+∠ORQ)
        ∠POQ=2∠PRQ
        Case III: When arc(PQ) is a major arc.
        We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles
        In ΔPOR, we have
        ∠POM=∠ORP+∠ORP           [OP=OR=r, ∠OPR=∠ORP]
        ⇒ ∠POM=2∠ORP ……………….(v)
        In ΔQOR, we have
        ∠QOM=∠ORQ+∠OQR
        ⇒ ∠QOM=2∠ORQ ……………….(vi)
        Adding equations (v) and (vi), we get
        ∠POM+∠QOM=2(∠ORP+∠ORP)
        ⇒ Reflex ∠POQ=2∠PRQ

         

        (19)Prove that Angles in the same segment of a circle are equal.

        Given: A circle C(O,r), an arc PQ and two angles ∠PRQ and ∠PSQ in the same segment of  the circle.
        To Prove: ∠PRQ= ∠PSQ
        Construction: Join OP and OQProof: we know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point in the remaining part of the circle. So we have
        ∠POQ=2∠PRQ and ∠POQ=2∠PSQ
        ⇒ 2∠PRQ=2∠PSQ
        ⇒ ∠PRQ=∠PSQ
        We have
        Reflex ∠POQ=2∠PRQ and ∠POQ=2∠PSQ
        ⇒ 2∠PRQ=2∠PSQ
        ⇒ ∠PRQ=∠PSQ
        Thus , in both the cases, we have
        ∠PRQ=∠PSQ

        (20) Prove that The angle in a semi-circle is a right angle.

        Given: PQ is a diameter of a circle C(O,r) and ∠PRQ is an angle in semi-circle.
        To Prove: ∠POQ=90∘Proof: we know that the angle subtended by an arc of a circle at its centre is twice the angle formed by the same arc at a point on the circle. So, we have
        ∠POQ=∠PRQ
        ⇒ 180∘=2∠PRQ                                    [POQ is a straight line]
        ⇒ ∠PRQ=90∘

        (21) Prove that The opposite angles of a cyclic quadrilateral are supplementary.

        Given: A Cyclic quadrilateral ABCD
        To Prove: ∠A+∠C=180∘ and ∠B+∠D=180∘
        Construction: Join AC and BD.

        Proof: Consider side AB of quadrilateral ABCD as the Chord of the circle. Clearly, ∠ACB and ∠ADB are angles in the same segment determined by chord AB of the Circle.
        ∠ACB=∠ADB          ………….(i)
        Now , consider the side BC of quadrilateral ABCD as the chord of the circle. We find that ∠BAC  and ∠BDC are angles in the same segment
        ∠BAC  = ∠BDC             [angles in the same segment are equal]..(ii)
        Adding equation (i) and (ii), we get
        ⇒       ∠ACB+∠BAC=∠ADB+∠BDC
        ⇒       ∠ACB+∠BAC=∠ADC
        ⇒       ∠ABC+∠ACB+∠BAC=∠ABC+∠ADC
        ⇒       180∘=∠ABC+∠ADC         [sum of angle of triangle is 180∘ ]
        ⇒       ∠ABC+∠ADC=180∘
        ⇒       ∠B+∠D=180∘
        But, ∠A+∠B+∠C+∠D=360∘
        ∠A+∠C=360∘−(∠B+∠D)
        ⇒ ∠A+∠C=360∘−180∘=180∘
        Hence, ∠A+∠C=180∘ and ∠B+∠D=180∘
        The converse of this theorem is also true as given below.

        (22) Prove that If the sum of any pair of opposite angles of a quadrilateral is 180∘,then it is cyclic.

        Given: A quadrilateral ABCD in which ∠B+∠D=180∘
        To Prove: ABCD is acyclic quadrilateral.Proof: If possible, Let ABCD be not cyclic quadrilateral. Draw a circle passing through three non-collinear points A, B and C. Suppose the circle meets AD or AD produced at D′. Join D′C.
        Now, ABCD’ is a cyclic quadrilateral.
        ∠ABC+∠AD′C=180∘…………..(i)
        But, ∠B+∠D=180∘
        i.e. ∠ABC+∠ADC=180∘…………..(ii)
        from (i) and (ii), we get
        ∠ABC+∠AD′C = ∠ABC+∠ADC
        ⇒       ∠AD′C = ∠ADC
        ⇒       An exterior angle of ΔCDD′ is equal to interior oppsite angle.
        But, this is not possible, unless D′ coincides with D. Thus, the circle passing through A,B,C also passes through D.
        Hence, ABCD is a cyclic Quadrilateral.

        (23) Prove that If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.

        Given: A Cyclic quadrilateral ABCD one of whose side AB is produced to E.
        To Prove: ∠CBE=∠ADCProof: Since ABCD is a quadrilateral and the sum of opposite pairs of angles in a cyclic quadrilateral is 180∘
        ∠ABC+∠ADC=180∘
        But,     ∠ABC+∠CBE=180∘                                  [Liner Pairs]
        ∠ABC+∠ADC=∠ABC+∠CBE
        ⇒       ∠ADC=∠CBE
        Or,                               ∠CBE=∠ADC

        (24) Prove that The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.

        Given: A Cyclic quadrilateral ABCD in which AP,BP,CR and DR are the bisectors of ∠A, ∠B, ∠C and ∠D respectively such that a quadrilateral PQRS is formed.
        To Prove: PQRS is a cyclic quadrilateral.Proof: In order to prove that PQRS is a cyclic quadrilateral, it is sufficinet to show that
        ∠APB+∠CRD=180∘
        Since the sum of the angles of a triangle is 180∘. Therefore, in triangles APB and CRD, we have
        ∠APB+∠PAB+∠PBA=180∘
        And,                ∠CRD+∠RCD+∠RDC=180∘
        ⇒       ∠APB+12∠A+12∠B=180∘
        And,                ∠CRD+12∠C+12∠D=180∘
        ⇒       ∠APB+12∠A+12∠B+∠CRD+12∠C+12∠D=180∘+180∘
        ∠APB+∠CRD+12{∠A+∠B+∠C+∠D}=360∘
        ∠APB+∠CRD+12{(∠A+∠C)+(∠B+∠D)}=360∘
        ∠APB+∠CRD+12(180∘+180∘)=360∘
        ∠APB+∠CRD=180∘
        Hence, PQRS is a cyclic Quadrilateral.

         

        (25) Prove that If two sides cyclic quadrilateral are parallel, then the remaining two sides are equal and the diagonals are also equal.

        Given: A Cyclic quadrilateral ABCD in which AB∥DC.
        To Prove: (i) AD=BC          (ii) AC=BDProof: In order to prove the desired results, it is sufficient to show that ΔADC≅ΔBCD. Since ABCD is cyclic Quadrilateral and sum of oppsite pairs of angles in a cyclic Quadrilateral is 180∘
        ∠B+∠D=180∘……..(i)
        Since AB∥DC and BC is a transveral and sum of the interior angles on the same side of a transversal is 180∘
        ∠ABC+∠BCD=180∘
        ∠B+∠C=180∘……………….(ii)
        From (i) and (ii), we get
        ∠B+∠D=∠B+∠C
        ⇒       ∠C=∠D……………..(iii)
        Now, consider triangles ADC and BCD. In ΔADC and ΔBCD, we have
        ∠ADC=∠BCD           [From equation (iii)]
        DC=DC                                  [Common]
        And,                ∠DAC=∠CBD           [∠DAC and ∠CBD are angles in the segment of chord CD]
        So, by AAS-criterion of congruence, we have
        ΔADC≅ΔBCD
        ⇒       AD=BC and AC=BD

        (26) Prove that If two opposite sides of a cyclic quadrilateral are equal, then the other two sides are parallel.

        Given: A cyclic quadrilateral ABCD such that AD=BC.
        To Prove: AB∥CD
        Construction: Join BD.Proof: We have,
        AD=BC
        ⇒       DA⌢≅BC⌢
        ⇒       m(DA)⌢≅(BC)⌢
        ⇒       2∠2=2∠1
        ⇒       ∠2=∠1
        But, these are alternate interior angles. Therefore, AB∥CD.

        (27) Prove that An isosceles trapezium is cyclic.

        Given: A trapezium ABCD in which AB∥DC and AD=BC
        To Prove: ABCD is a cyclic trapezium.
        Construction: Draw DE⊥AB and CF⊥AB.Proof: In order to prove that ABCD is a cyclic trapezium, it is sufficient to show that ∠B+∠D=180∘.
        In triangles DEA and CFB, we have
        AD=BC                        [Given]
        ∠DEA=∠CFB           [Each equal to 90∘]
        And,                DE=CF
        So, by RHS-criterion of congruence, we have
        ΔDEA≅ΔCFB
        ⇒       ∠A=∠B and ∠ADE=∠BCF
        Now,               ∠ADE=∠BCF
        ⇒       90∘+∠ADE=90∘+∠BCF
        ⇒       ∠EDC+∠ADE=∠FCD+∠BCF
        ⇒       ∠ADC=∠BCD
        ⇒       ∠D=∠C
        Thus,               ∠A=∠B and ∠C=∠D.
        ∠A+∠B+∠C+∠D=360∘
        ⇒       2∠B+2∠D=360∘
        ⇒       ∠B+∠D=180∘
        Hence, ABCD is a cyclic quadrilateral.

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