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BOOKWORK: EXAMINABLE PROOFSS
PAPER 2: GEOMETRY
Circle Geometry Theorems
These are the
A
The line segment drawn from the centre of a circle,
1
examinable proofs perpendicular to a chord, bisects the chord.
T
Given : ?O with OP AB
required for Grade 12. To prove : AP = PB
Construction : Join OA and OB O
s
Proof : In Δ OPA & OPB
(1) OA = OB . . . radii 1 2
5 Grade 11 &
A B
P
(2) Pˆ 1 = Pˆ 2 (= 90º) . . . given
(3) OP is common
â ΔOAP ≡ ΔOBP . . . RHS
2 Grade 12 â AP = PB, i.e. OP bisects chord AB
This proof has
The line drawn from the centre of a circle that ...
2 been added in the
bisects a chord is perpendicular to the chord.
2021 Exam Guidelines.
BOOKWORK: EXAMINABLE PROOFS
Given : ?O with AP = PB
To prove : OP AB
Construction : Join OA and OB
O
Proof : In Δs OPA & OPB
(1) OA = OB . . . radii 1 2
A B
P
(2) AP = PB . . . given
(3) OP is common
â OAP ≡ ΔOBP . . . SSS
Pˆ 1 = Pˆ 2
But, Pˆ 1 + Pˆ 2 = 180º . . . øs on a straight line
â Pˆ 1 = Pˆ 2 = 90º
i.e. OP AB
Copyright © The Answer Series: Photocopying of this material is illegal i
, The angle which an arc of a circle subtends at the centre is The angle between a tangent to a circle and a chord drawn from the point of
3 5
double the angle it subtends at any point on the circumference. contact is equal to the angle subtended by the chord in the alternate segment.
Given : ˆ at the centre and APB
?O, arc AB subtending AOB ˆ at the circumference.
Method 1
S
To prove : ˆ = 2APB
AOB ˆ Draw radii and use
P
'ø at centre' theorem.
M
Construction : Join PO and produce it to Q.
12 Given : ?O with tangent at N and chord NM
A
x y subtending K̂ at the circumference.
Proof : Let Pˆ 1 = x K O
Q
O RTP : ˆ = K̂
MNQ
s
Then  = x . . . ø opposite equal radii x 1 2
Construction : radii OM and ON
T
A y
â Oˆ 1 = 2x . . . exterior ø of ΔAOP N
Q
Proof : ˆ = x
Let MNQ
B
Similarly, if Pˆ 2 = y, then Oˆ 2 = 2y ˆ = 90º . . . radius tangent
ONQ P
ˆ
â AOB = 2 x + 2y ˆ = 90º – x
ONM
= 2( x + y) ˆ = 90º – x . . . øs opposite equal radii
OMN
ˆ
= 2 APB ˆ = 2x . . . sum of øs in
MON
K̂ = x . . . ø at centre = 2 % ø at circumference
ˆ
â MNQ = K̂
4 The opposite angles of a cyclic quadrilateral are supplementary.
These proofs
Given : ?O and cyclic quadrilateral ABCD A Method 2 are logical &
x easy to follow.
To prove : Â + Ĉ = 180º & B̂ + D̂ = 180º We use 2 'previous' facts involving right øs
2 tangent diameter . . . so, draw a diameter !
Construction : Join BO and DO. O D
1 ø in semi-? = 90º . . . so, join RK !
BOOKWORK: EXAMINABLE PROOFS
Proof : Let  = x B
Given : ?O with tangent at N and chord NM
ˆ = 2x ø at centre = 2 % C
Then O1 ... subtending K̂ at the circumference. R
ø at circumference
ˆ = MKN
ˆ M
â Oˆ 2 = 360º – 2x . . . øs about point O RT P : MNQ
ø at centre = 2 %
â Ĉ = 1 (360º – 2x) = 180º – x ... Construction : diameter NR; join RK
K O
2 ø at circumference Q
Proof : ˆ
RNQ = 90º . . . tangent diameter
â Â + Ĉ = x + 180º – x = 180º
ˆ
& RKN = 90º . . . ø in semi-? x
& â B̂ + D̂ = 180º ... sum of the øs of a
quadrilateral = 360º Then . . . N
Let ˆ
MNQ = x
ˆ P
â RNM = 90º – x
ˆ
â RKM = 90º – x . . . øs in same segment
ˆ = x
â MKN
ˆ = MKN
â MNQ ˆ
ii Copyright © The Answer Series: Photocopying of this material is illegal
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