Method One<br />
Let Angle AOB be x°.<br />We draw a line from O to P. <br />Since lines OP, OB and OA are radii of the same circle, they a...
Let Angle BOP be y°.<br />Angle OBP = Angle OPB = (180°-y°)/2<br />------------- Sum of interior angles of a triangle is 1...
Angle AOP = x°+y°<br />Angle OAP = Angle OPA = (180°-x°-y°)/2<br />------------- Sum of interior angles of a triangle is 1...
Angle APB = Angle OPB – Angle OPA<br />	     = [(180°-y°)/2] – [(180°-x°-y°)/2] <br />	     = (180°-y°-180°+x°+y°)/2 <br /...
Method TWO<br />
We draw a line form P to O and extend it out to C, forming the diameter of the circle.<br />Therefore OC=OC=OB=OP since th...
Let angle OBP be of value x°.<br />Since triangle OBP is an isosceles triangle, angle OBP=angle OPB=x°<br />Therefore angl...
Let angle OAP be of value y°.<br />Since triangle OAP is an isosceles triangle, angle OAP=angle OPA=y°<br />Therefore angl...
Angle AOB= angle COB-angle COA<br />	     = 2x°-2y°<br />	     = 2(x-y)°<br />Angle APB= angle OPB-angle OPA<br />	   = x°...
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Part A) Proof

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MA301 Assignment: Proof for property of circle

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Part A) Proof

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

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