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Prove that the angle at the circumference is always 90 0

Prove that the angle at the circumference is always 90 0. P. α. That is, no matter where you place point P, the angle α is always 90 0. A. B. O. Note: AB is the diameter of the circle whose centre is at O. P. Prove that the angle at the circumference is always 90 0.

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Prove that the angle at the circumference is always 90 0

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  1. Prove that the angle at the circumference is always 900 P α That is, no matter where you place point P, the angle α is always 900 A B O Note: AB is the diameter of the circle whose centre is at O

  2. P Prove that the angle at the circumference is always 900 Mark in the radius OP y x 2 isosceles triangles are thus formed y x A B So we can mark in angles x and y O Now we add the angles in Triangle APB The angle sum is x + x + y + y = 2x + 2y So 2x + 2y = 1800 x + y = 900 So angle APB is always 900

  3. Prove that the angle at the centre of the circle is always double the angle at the circumference P b That is, no matter where you place point P, the angle α = 2b always O a B AB is a chord of the circle with centre O A

  4. P Mark in the radius OP Prove the angle at the centre is always double the angle at the circumference 3 isosceles triangles are thus formed z y So we can mark in angles x, y and z O Angle APB = y + z y z Angle AOB = 1800 - 2x x B x A But, looking at triangle APB, 2x + 2y + 2z = 1800  1800 – 2x = 2(y + z) Angle AOB = 2 x Angle APB

  5. Prove that angles in the same segment are always equal That is, no matter where you place point P, the angle APB always the same – that is a = b P1 P2 a b AB is a chord of the circle, which splits the circle into 2 segments B A

  6. P2 Mark in the centre O, and form the triangle AOB Prove that angles in the same segment are equal P1 a a Let the angle AOB be 2a O Thus angle AP1B will be a (angle at centre = double angle at circumference) 2a B A Angle AP2B will be a for the same reason Thus angle AP1B = Angle AP2B

  7. Prove that equal chords are equidistant from the centre of a circle C That is, if AB = CD, then d1 = d2 d2 D O d1 AB and CD are chords of equal length B A

  8. Prove that equal chords are equidistant from the centre of a circle C P is the mid-point of AB, and Q is the mid-point of CD Q AP = CQ D O Mark in OA and OC (both are radii) Triangle OPA is congruent to triangle OQC B P A  OP = OQ

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