## Angles on a Circle

Part of the Geometry Primer for Mathematics 337 at the University of British Columbia.

 Let us reconsider the argument which proves that a right angle is formed whenever we connect any point P on a semi-circle with the two end points A and B of its base: the green (blue) angle at P equals half of the red (yellow) angle at centre C of the circle. Hence the green and blue angles at P add up to one half of the sum of the red and yellow angles (which is 180 degrees).

 Except for its last parenthesis, this argument carries over verbatim to segments AB which are not necessarily diameters of the circle -- as long as the point P does not wander off too far to the left or the right; more precisely, as long as C still lies inside the triangle ABP. Upshot: the angle APB at the circumference is exactly half of the angle ACB at the centre. In other words, keeping A and B fixed, you can move P around without changing the angle at the top. This unexpected result would be an even nicer theorem without that awkward restriction on the movement of P.
 In fact, the desired theorem does hold! However, if C lies outside the triangle ABP, the picture is quite different -- but if you switch your thinking from adding to subtracting, you will see a similar logical pattern. It must be shown that the yellow angle at C equals twice the blue one at P. The latter is the difference between the blue one at B and the yellow one at A -- base angles of the isosceles triangles PCB and PCA. The exterior angles at C of the same triangles are twice as big. They are the red and yellow angle for PCB and the thin red one for PCA -- and their difference is just the yellow angle at C.

 What about a point Q on the other side of AB, where the circular segment is less than a semi-circle? This is easily answered in terms of the point P diametrically opposite Q: the angle AQB is just the supplement of our old friend BPA. Indeed the green and the blue triangles shown here have right angles at A and B, respectively, hence their angles at P and Q add up to 90 degrees. The total angles (green and blue together) at P and Q therefore add to 180 degrees. Summary: all angles on one side of a circular segment are equal, and are supplementary to the angles on the other side of that segment.
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