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Proof for angle property of circle. LeowKeXuan, Ng Ling Yan, Tan Yu Xuan, Cheong Jing Fu, Yang Le.
So as we all knowโ€ฆ Triangle COB is an isosceles triangle (radius of circle = equal length), so if angle OCB is x, angle OCB is also x. Hence, angle DOB,  being the exterior angle of triangle COB, =2x. This also applies for triangle COA. Therefore, angle ACB = x + y, angle AOB = 2(x + y).  This proves that the angle subtended by an arc at the centre of a circle is twice that subtended at the circumference. Source: Math Textbook 3 Page 370. C x y O 2y 2 x y x B A D
But in this caseโ€ฆ P O . B A The same prove canโ€™t be use!
Thereforeโ€ฆ We came up with an alternative prove which can be used in this case!  Extend the line OB to touch the circumference of the circle and mark the new point as C.  We know that triangle COA is an isosceles triangle as CO = AO (radius of circle.) Therefore, angle OCA = angle OAC = angle x. Angle APB also = angle x (angles in the same segment) Angle AOB = 2x (exterior angle of a triangle) Angle AOB (2x) is equivalent to 2 of angle APB (x) HENCE PROVEN!  C x P O x . x B A
THE END! :D

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Circle.

  • 1. Proof for angle property of circle. LeowKeXuan, Ng Ling Yan, Tan Yu Xuan, Cheong Jing Fu, Yang Le.
  • 2. So as we all knowโ€ฆ Triangle COB is an isosceles triangle (radius of circle = equal length), so if angle OCB is x, angle OCB is also x. Hence, angle DOB, being the exterior angle of triangle COB, =2x. This also applies for triangle COA. Therefore, angle ACB = x + y, angle AOB = 2(x + y). This proves that the angle subtended by an arc at the centre of a circle is twice that subtended at the circumference. Source: Math Textbook 3 Page 370. C x y O 2y 2 x y x B A D
  • 3. But in this caseโ€ฆ P O . B A The same prove canโ€™t be use!
  • 4. Thereforeโ€ฆ We came up with an alternative prove which can be used in this case! Extend the line OB to touch the circumference of the circle and mark the new point as C. We know that triangle COA is an isosceles triangle as CO = AO (radius of circle.) Therefore, angle OCA = angle OAC = angle x. Angle APB also = angle x (angles in the same segment) Angle AOB = 2x (exterior angle of a triangle) Angle AOB (2x) is equivalent to 2 of angle APB (x) HENCE PROVEN! C x P O x . x B A