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You can prove this just from knowing that the three angles of a triangle add to 180 degrees.

Label the ends of the arc A and B, the circle's centre O and the "any other point" P.

We want to show that angle APB is half of angle AOB.

(This uses the notation where you write three letters XYZ to mean the angle formed between straight lines XY and YZ).

Consider the triangle formed by O, B and P. The lengths OB and OP are the same (the radius of the circle). This means that the triangle is isosceles and angles OPB and OBP are the same.

Since POB + OPB + OBP = 180 degrees

and OPB = OBP

we have POB + 2 * OPB = 180

With the same working, the triangle formed by O, A and P gives us

POA + 2 * OPA = 180

Subtract these two equations:

(POB - POA) + 2 * (OPB - OPA) = 180 - 180 = 0

Rearrange:

2 * (OPB - OPA) = POA - POB

If you draw a diagram you will see that OPB - OPA = APB

and POA - POB = AOB

so we have 2 * APB = AOB as required.

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Q: Angle subtended by an arc is double the angle subtended by it at any other point on the remaining part of the circle?
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