Page 130 - Mathematics_Form_3

Circles

Case IV In case I, given a circle centred at O
X ˆ
with arc AB subtending angle AOB at
ˆ
the centre and AXB at circumference.
B
ˆ
To prove AOB 2AXB= ˆ .
Proceed as follows:
FOR ONLINE READING ONLY
A O
Construction: Join XO extended to Y.
Proof: In AOX∆ , it implies that

AO = OX (radii of the same
circle)
ˆ
AXO OAX= ˆ (base angles of

Theorem 5.1 can be proved using the isosceles triangle)
concepts of congruence, geometry and
ˆ
ˆ
algebra. AOY = 2AXO (exterior angle of

triangle) (1)
Proof of Theorem 5.1 for Cases I
and III Similarly, in BOX∆ , it follows that
BO = OX (radii of the same
Case I
X circle)
ˆ
ˆ
BXO = XBO(base angles of

isosceles triangle)
ˆ
ˆ

O BOY = 2BXO (exterior angle of
triangle) (2)
ˆ
But, AOB AOY BOY= ˆ + ˆ
A B
From equations (1) and (2), it implies
Y that
ˆ
Case III AOB 2AXO 2BXO= ˆ + ˆ
ˆ
X AOB 2 AXO BXO= ( ˆ + ˆ ) Mathematics for Secondary Schools
ˆ
AOB 2AXB= ˆ
For Case III, consider OXA∆ .
O
AO = XO (radii of the same circle)
ˆ
ˆ
OAX = OXA (base angles of
A isosceles triangle)
ˆ
B AOB OAX OXA= ˆ + ˆ (exterior angle
of the triangle OXA )

Student\s Book Form Three 123

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MATHEMATIC F3 SB.indd 123 18/09/2025 09:59:38
MATHEMATIC F3 SB.indd 123