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Proof. Circle Theorems. Contents. Angle at Centre Opposite angles in cyclic quadrilateral Alternate Segment Theorem. Circle Theorems on the Web. O. R. P. Prove angle at centre is twice the angle at the circumference. Q. Draw line QO and extend to intersect circumference at S. a. b.
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Proof Circle Theorems
Contents • Angle at Centre • Opposite angles in cyclic quadrilateral • Alternate Segment Theorem Circle Theorems on the Web
O R P Prove angle at centre is twice the angle at the circumference Q Draw line QO and extend to intersect circumference at S a b Label angles <OQR is a <OQP is b S
O R P S Prove angle at centre is twice the angle at the circumference Q <ORQ = <OQR Base angles in Isosceles Triangle a b (OQ = OR, radii of circle) Similarly <OPQ = <OQP a b
O R P S Prove angle at centre is twice the angle at the circumference Q Consider <ROS a By exterior angle of triangle theorem b <ROS = 2a Similarly 2a 2b a b <POS = 2b
O R P S Prove angle at centre is twice the angle at the circumference Q <POR = 2a + 2b = 2(a + b) a (a+b) b <PQR = a + b <POR = 2 x <PQR 2a 2b 2(a+b) Return to Menu
Opposite Angles in a Cyclic Quadrilateral are supplementary Required to Prove that x + y = 180º x Draw in radii The angle at the centre is TWICE the angle at the circumference 2y 2x 2x + 2y = 360º y 2(x + y) = 360º x + y = 180º Opposite Angles in Cyclic Quadrilateral are Supplementary Return to Menu
Alternate Segment Theorem R Given that Q MPN is a Tangent to the circle S PS is a chord in the circle Required to Prove … that < SPN = Any angle in the major segment PQS M P N Draw diameter PR Join RS
Alternate Segment Theorem R (Angle in Semicircle) <PSR = 90º Q (Radius to a Tangent) <NPR = 90º <NPS + <RPS = 90º S <PRS + <RPS = 90º <PRS + <RPS = <NPS + <RPS <RPS is common M P N <PRS = <NPS Special case of Rightangle Triangle on circle’s diameter
Alternate Segment Theorem R Now for ANY angle in Segment PQS Q Join QS & PQ S Consider <PQS <PQS = <PRS (Angles in SAME segment) M P N Thus <SPN = Any Angle in Alternate Segment Return to Menu