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)
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)°].
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