(a) Proving Angle Properties of Circles<br />
In Geometry, there is a property which states <br />that the angle subtended by an arc at the centre of a circle is twice ...
But what you were given this figure where it is impossible to draw a straight line that passes through the centre O and st...
Let us show you how to :D<br />
1<br />Extend line AO<br />C<br />Draw a line from O to P<br />2<br />O<br />P<br />B<br />A<br />
∠AOB + ∠BOP + ∠POC <br />= 180°<br />(adj. angles on a straight line)<br />C<br />O<br />P<br />∠OBP = ∠OPB<br />(isoscele...
∠AOB + 180 ° – 2 x ∠OPB +∠POC =180 °<br />C<br />∠OPB = ∠OPA +∠APB<br />O<br />P<br />O<br />P<br />B<br />A<br />B<br />A...
∠POC = 2∠OPA<br />[∠OPA = ∠OAP<br />(isosceles triangle),<br />∠POC = ∠OAP + ∠OPA<br />(external angles)]<br />C<br />C<br...
Therefore…<br />SOLVING THE EQUATION GIVE US<br />C<br />∠AOB = 2 ∠APB<br />O<br />P<br />PROVEN!! :D<br />B<br />A<br />
The end :D<br />
Done by<br />Ming Xuan<br />Cheng Mun<br />Natalie<br />Jerald<br />Xin Yao<br />Xin Yi<br />
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A) proving angle properties of circles 2

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A) proving angle properties of circles 2

  1. 1. (a) Proving Angle Properties of Circles<br />
  2. 2. In Geometry, there is a property which states <br />that the angle subtended by an arc at the centre of a circle is twice that subtended at the circumference of the same circle.<br />P<br />O<br />However for this proof, it is assumed that we are always able to draw a straight line (i.e. POC) which passes through the centre O, and cuts both ∠AOB and ∠APB.<br />∠AOB = 2 x ∠APB<br />B<br />A<br />e.g. If ∠AOB = 50°, <br />then according to the property, ∠APB = 25°.<br />C<br />
  3. 3. But what you were given this figure where it is impossible to draw a straight line that passes through the centre O and still can cut both ∠AOB and ∠APB?<br />O<br />P<br />B<br />A<br />How do you proof it then?<br />
  4. 4. Let us show you how to :D<br />
  5. 5. 1<br />Extend line AO<br />C<br />Draw a line from O to P<br />2<br />O<br />P<br />B<br />A<br />
  6. 6. ∠AOB + ∠BOP + ∠POC <br />= 180°<br />(adj. angles on a straight line)<br />C<br />O<br />P<br />∠OBP = ∠OPB<br />(isosceles triangle) <br />B<br />A<br />∠BOP = 180 ° – 2 x ∠OPB <br />O<br />P<br />
  7. 7. ∠AOB + 180 ° – 2 x ∠OPB +∠POC =180 °<br />C<br />∠OPB = ∠OPA +∠APB<br />O<br />P<br />O<br />P<br />B<br />A<br />B<br />A<br />∠AOB – 2(∠OPA + ∠APB) + ∠POC = O<br />∠AOB – 2∠OPA – 2∠APB + ∠POC = O<br />
  8. 8. ∠POC = 2∠OPA<br />[∠OPA = ∠OAP<br />(isosceles triangle),<br />∠POC = ∠OAP + ∠OPA<br />(external angles)]<br />C<br />C<br />O<br />P<br />O<br />P<br />B<br />A<br />B<br />A<br />∠AOB - 2 ∠OPA - 2 ∠APB + ∠POC = O<br />
  9. 9. Therefore…<br />SOLVING THE EQUATION GIVE US<br />C<br />∠AOB = 2 ∠APB<br />O<br />P<br />PROVEN!! :D<br />B<br />A<br />
  10. 10. The end :D<br />
  11. 11. Done by<br />Ming Xuan<br />Cheng Mun<br />Natalie<br />Jerald<br />Xin Yao<br />Xin Yi<br />

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