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Proof

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Base angles in Isosceles Triangle. Similarly OPQ = OQP (OQ = OR, radii of circle) ... Special case of Rightangle Triangle on circle's diameter. Alternate ... – PowerPoint PPT presentation

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Title: Proof


1
Proof
  • Circle Theorems

2
Contents
  • Angle at Centre
  • Opposite angles in cyclic quadrilateral
  • Alternate Segment Theorem

Circle Theorems on the Web
3
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
ltOQR is a
ltOQP is b
S
4
Prove angle at centre is twice the angle at the
circumference
Q
ltORQ ltOQR
Base angles in Isosceles Triangle
a
b
(OQ OR, radii of circle)
Similarly ltOPQ ltOQP
a
b
5
Prove angle at centre is twice the angle at the
circumference
Q
Consider ltROS
a
By exterior angle of triangle theorem
b
ltROS 2a
Similarly
2a
2b
a
b
ltPOS 2b
6
Prove angle at centre is twice the angle at the
circumference
Q
ltPOR 2a 2b
2(a b)
a
(ab)
b
ltPQR a b
ltPOR 2 x ltPQR
2a
2b
2(ab)
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7
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
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8
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 lt SPN Any angle in the major segment PQS
M
P
N
Draw diameter PR
Join RS
9
Alternate Segment Theorem
R
(Angle in Semicircle)
ltPSR 90º
Q
(Radius to a Tangent)
ltNPR 90º
S
ltNPS ltRPS 90º
ltPRS ltRPS 90º
ltPRS ltRPS ltNPS ltRPS
ltRPS is common
M
P
N
ltPRS ltNPS
Special case of Rightangle Triangle on circles
diameter
10
Alternate Segment Theorem
R
Now for ANY angle in Segment PQS
Q
Join QS PQ
S
Consider ltPQS
ltPQS ltPRS
(Angles in SAME segment)
M
P
N
Thus ltSPN Any Angle in Alternate Segment
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