text: # Theorem – The sum of opposite angles of a cyclic quadrilateral is 180° | Class 9 Maths
  
  • Difficulty Level : Medium
  • Last Updated : 12 Oct, 2021
  
  In Euclidean geometry, a cyclic quadrilateral or inscribed quadrilateral is a quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic. The centre of the circle and its radius are called the circumcenter and the circumradius respectively. Other names for these quadrilaterals are concyclic quadrilateral and chordal quadrilateral, the latter since the sides of the quadrilateral are chords of the circumcircle.
  
  ### Theorem Statement: The sum of the opposite angles of a cyclic quadrilateral is 180°.
  
  So according to the theorem statement, in the below figure, we have to prove that
  
  ∠BAD + ∠BCD = 180o
  
  ∠ABC + ∠ADC = 180o
  
  Proof:
  
  Given:
  
  A cyclic quadrilateral ABCD where O is the centre of a circle.
  
  Construction:
  
  Join the line segment OB and OD
  
  Since, The angle subtended by an arc at the centre is double the angle on the circle.
  
  Therefore,
  
  ∠BAD = 1/2 (∠BOD)                         – equation 1
  
  Similarly,
  
  ∠BCD = 1/2 (reflex of ∠BOD)           – equation 2
  
  By adding equation 1 and 2, we get
  
  ∠BAD + ∠BCD = 1/2 (∠BOD + reflex of ∠BOD)
  
  ∠BAD + ∠BCD = 1/2 ( 360o)             {Since , ∠BOD + reflex of ∠BOD form complete angle i.e 360 degree}
  
  ∠BAD + ∠BCD = 180o
  
  Similarly,
  
  ∠ABC + ∠ADC = 180o
  
  Hence proved, that the sum of opposite angles of a cyclic quadrilateral is 180°
  
  ### Sample Problems on the Given Theorem
  
  Question 1: In the figure given below, ABCD is a cyclic quadrilateral in which ∠CBA = 91.64° and ∠DAB = 102.51° find ∠ADC and ∠DCB?
  
  Solution:
  
  By applying theorem,
  
  The sum of opposite angles of a cyclic quadrilateral is 180°
  
  We get,
  
  ∠ADC + ∠ABC = 180o
  
  ∠ADC + 91.64o =180o
  
  ∠ADC =180o – 91.64o = 88.36 degree
  
  And,
  
  ∠DCB + ∠BAD =180o
  
  ∠DCB + 102.51o = 180o
  
  ∠DCB = 180o – 102.51o
  
  ∠DCB = 77.49o
  
  Hence ∠ADC = 88.36o and ∠DCB = 77.49o
  
  Question 2: In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 100° and ∠BDC = 50° find ∠DBC?
  
  Solution:
  
  Given :
  
  ∠BAD = 100°  and ∠CDB = 50o
  
  ∠BAD + ∠BCD = 180  (Since opposite angle of cyclic quadrilateral)
  
  ∠BCD = 180o – 100o = 800
  
  In ∆BCD ,
  
  ∠BCD + ∠CDB + ∠DBC = 180o
  
  80o  + 50o + ∠DBC = 180o
  
  ∠DBC = 180o  – 130o
  
  ∠DBC = 50o
  
  Therefore ∠DBC = 50
  
  My Personal Notes arrow_drop_up
explanation: The document provides a detailed explanation of a theorem in Euclidean geometry, specifically about the sum of opposite angles of a cyclic quadrilateral being 180°. It includes a proof of the theorem and two sample problems with solutions. The sample problems contain questions and their corresponding answers.
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distilabel_metadata.content: # Theorem – The sum of opposite angles of a cyclic quadrilateral is 180° | Class 9 Maths
  
  • Difficulty Level : Medium
  • Last Updated : 12 Oct, 2021
  
  In Euclidean geometry, a cyclic quadrilateral or inscribed quadrilateral is a quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic. The centre of the circle and its radius are called the circumcenter and the circumradius respectively. Other names for these quadrilaterals are concyclic quadrilateral and chordal quadrilateral, the latter since the sides of the quadrilateral are chords of the circumcircle.
  
  ### Theorem Statement: The sum of the opposite angles of a cyclic quadrilateral is 180°.
  
  So according to the theorem statement, in the below figure, we have to prove that
  
  ∠BAD + ∠BCD = 180o
  
  ∠ABC + ∠ADC = 180o
  
  Proof:
  
  Given:
  
  A cyclic quadrilateral ABCD where O is the centre of a circle.
  
  Construction:
  
  Join the line segment OB and OD
  
  Since, The angle subtended by an arc at the centre is double the angle on the circle.
  
  Therefore,
  
  ∠BAD = 1/2 (∠BOD)                         – equation 1
  
  Similarly,
  
  ∠BCD = 1/2 (reflex of ∠BOD)           – equation 2
  
  By adding equation 1 and 2, we get
  
  ∠BAD + ∠BCD = 1/2 (∠BOD + reflex of ∠BOD)
  
  ∠BAD + ∠BCD = 1/2 ( 360o)             {Since , ∠BOD + reflex of ∠BOD form complete angle i.e 360 degree}
  
  ∠BAD + ∠BCD = 180o
  
  Similarly,
  
  ∠ABC + ∠ADC = 180o
  
  Hence proved, that the sum of opposite angles of a cyclic quadrilateral is 180°
  
  ### Sample Problems on the Given Theorem
  
  Question 1: In the figure given below, ABCD is a cyclic quadrilateral in which ∠CBA = 91.64° and ∠DAB = 102.51° find ∠ADC and ∠DCB?
  
  Solution:
  
  By applying theorem,
  
  The sum of opposite angles of a cyclic quadrilateral is 180°
  
  We get,
  
  ∠ADC + ∠ABC = 180o
  
  ∠ADC + 91.64o =180o
  
  ∠ADC =180o – 91.64o = 88.36 degree
  
  And,
  
  ∠DCB + ∠BAD =180o
  
  ∠DCB + 102.51o = 180o
  
  ∠DCB = 180o – 102.51o
  
  ∠DCB = 77.49o
  
  Hence ∠ADC = 88.36o and ∠DCB = 77.49o
  
  Question 2: In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 100° and ∠BDC = 50° find ∠DBC?
  
  Solution:
  
  Given :
  
  ∠BAD = 100°  and ∠CDB = 50o
  
  ∠BAD + ∠BCD = 180  (Since opposite angle of cyclic quadrilateral)
  
  ∠BCD = 180o – 100o = 800
  
  In ∆BCD ,
  
  ∠BCD + ∠CDB + ∠DBC = 180o
  
  80o  + 50o + ∠DBC = 180o
  
  ∠DBC = 180o  – 130o
  
  ∠DBC = 50o
  
  Therefore ∠DBC = 50
  
  My Personal Notes arrow_drop_up
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document provides a detailed explanation of a theorem in Euclidean geometry, specifically about the sum of opposite angles of a cyclic quadrilateral being 180°. It includes a proof of the theorem and two sample problems with solutions. The sample problems contain questions and their corresponding answers.
  
  Label: multiple-question-and-answer