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Theorem 9.9 - Chapter 9 Class 9 Circles
Last updated at May 29, 2023 by Teachoo
Theorems
Theorem 9.1
Theorem 9.2 Important
Theorem 9.3 Important
Theorem 9.4
Theorem 9.5 Important
Theorem 9.6
Theorem 9.7 Important
Theorem 9.8
Theorem 9.9 Important You are here
Theorem 9.10
Theorem 9.11 Important
Angle in a semicircle is a right angle Important
Only 1 circle passing through 3 non-collinear points Deleted for CBSE Board 2024 Exams
Transcript
Theorem 9.10 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). Given : A, B, C and D are 4 points (no 3 are collinear) AB subtends equal angles at C and D i.e. ∠ACB = ∠ADB. To Prove : A,B, C and D are concylic 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 Let circle intersect AD at D’ Now, ∠ACB = ∠AD’B But, given that ∠ACB = ∠ADB ∴ From (1) and (2) ∠ AD’B = ∠ ADB In ∆ BDD’ ∠ AD’B = ∠BDD’ + ∠D’ BD ∠ ADB = ∠ADB + ∠D’ BD ∠ ADB – ∠ADB = ∠D’ BD ∴ ∠D’BD = 0 ∴ D’ and D coincide Thus, Our assumption was wrong ⇒ Point D lies on circle ∴ A, B, C, D are concyclic. Hence proved
