Theorem 10.12 - Chapter 10 Class 9 Circles (Term 2)
Last updated at Aug. 25, 2021 by Teachoo
Transcript
Theorem 10.12 If the sum of a pair of opposite angles of a quadrilateral is 180 , the quadrilateral is cyclic. Given : ABCD is quadrilateral such that BAC + BDC = 180 Prove : ABCD is a cyclic quadrilateral Proof : Since A, B, C are non collinear One circle passes through three collinear points Let us draw a circle C1 with centre at O Let us assume D does not lie on C1 Now, ABCD is cyclic quadrilateral BAC + BD C = 180 But given BAC + BDC = 180 Thus, BD C = BDC Now, In BDD BD C = BDD + DBD BD C = BDC + DBD BDC = BDC + DBD BDC BDC = DBD DBD = 0 D and D Coincides Our assumption was wrong Point D lies on circle C1 A, B, C, D are concyclic. Hence proved
Theorems
Theorem 10.1
Theorem 10.2 Important
Theorem 10.3 Important
Theorem 10.4
Theorem 10.5 Deleted for CBSE Board 2022 Exams
Theorem 10.6 Important
Theorem 10.7
Theorem 10.8 Important
Theorem 10.9
Theorem 10.10 Important
Theorem 10.11
Theorem 10.12 Important You are here
Angle in a semicircle is a right angle Important
Chapter 10 Class 9 Circles (Term 2)
Serial order wise
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.