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# Part A) Proof

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MA301 Assignment: Proof for property of circle

MA301 Assignment: Proof for property of circle

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### Transcript

• 1. Method One
• 2. Let Angle AOB be x°.
We draw a line from O to P.
Since lines OP, OB and OA are radii of the same circle, they are equal in length.
From that, we can conclude that triangles OBP and OAP are isosceles triangles.
P
O
B
A
• 3. Let Angle BOP be y°.
Angle OBP = Angle OPB = (180°-y°)/2
------------- Sum of interior angles of a triangle is 180°
------------- Property of isosceles triangle
P
O
B
A
• 4. Angle AOP = x°+y°
Angle OAP = Angle OPA = (180°-x°-y°)/2
------------- Sum of interior angles of a triangle is 180°
------------- Property of isosceles triangle
P
O
B
A
• 5. Angle APB = Angle OPB – Angle OPA
= [(180°-y°)/2] – [(180°-x°-y°)/2]
= (180°-y°-180°+x°+y°)/2
= x°/2
P
O
B
A
Therefore, this proves that the angle at the centre (Angle AOB = x°) is twice the angle at circumference (Angle APB = x°/2)
• 6. Method TWO
• 7. We draw a line form P to O and extend it out to C, forming the diameter of the circle.
Therefore OC=OC=OB=OP since they are all radii of the circle.
Thus triangles OCA, OAP and OBP is isosceles.
P
O
C
B
A
• 8. Let angle OBP be of value x°.
Since triangle OBP is an isosceles triangle, angle OBP=angle OPB=x°
Therefore angle COB= 2x°.
------ Properties of exterior angles
P
O
C
B
A
• 9. Let angle OAP be of value y°.
Since triangle OAP is an isosceles triangle, angle OAP=angle OPA=y°
Therefore angle COA= 2y°.
------ Properties of exterior angles
P
O
C
C
B
A
• 10. Angle AOB= angle COB-angle COA
= 2x°-2y°
= 2(x-y)°
Angle APB= angle OPB-angle OPA
= x°- y°
= (x-y)°
P
O
C
B
A
Therefore, this proves that the angle at the centre [Angle AOB = 2(x-y)°] is twice the angle at circumference [Angle APB = (x-y)°].