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

Last updated at April 16, 2024 by Teachoo

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