Angles in a circle

[Sunflower124]

In this problem, I need to prove angle AOB = 2 angle ACB (2y). This is the answer's working. I understand up until it introduces a 2x in the third line. Where did this come from?
angle OBC = x+y (triangle OCB is isosceles)
angle COB = 180-2(x+y)
angle AOB = 180-2x-(180-2(x+y))
therefore angle AOB = 2y

 

[Deleted member 4993]

In this problem, I need to prove angle AOB = 2 angle ACB (2y). This is the answer's working. I understand up until it introduces a 2x in the third line. Where did this come from?
angle OBC = x+y (triangle OCB is isosceles)
angle COB = 180-2(x+y)
angle AOB = 180-2x-(180-2(x+y))
therefore angle AOB = 2yView attachment 27547

In the triangle OAC we have OC = OA

mOAC = m OCA = x

mAOC = 180 - 2 * x

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