Theorem 5.2: Let   P   be a partition of the arc between   A   and   B .   Then the inside approximation corresponding to P is shorter than the outside approximation.

If   Q   is a refinement of   P,   then the inside approximation corresponding to   Q   is larger than the inside approximation corresponding to   P,   and the outside approximation corresponding to   Q   is less than the outside approximation corresponding to   P.

Proof: By Theorem 5.1, the tangents will meet at points   T1, T2, . . . Tn,   so we can take the outside approximation to be

|AT1| + |T1P1| + |P1T2| + |T2P2| + . . . + |TnB|

The inside approximation is shorter than the outside approximation because since a straight line is the shortest distance between two points, by Theorem 3.5, the distance going straight from   Pi   to   Pi+1   will be shorter than the distance going through   Ti+1.

For the assertion concerning the refinement, it suffices to show that the result will hold if we add one extra point to the partition, because we could get from a partition to its refinement through a series of refinements where we add just one point at a time. Let   A   and   B   be two points in a partition, and let   C   be a point on the arc between   A   and   B.

|AB| < |AC| + |CD|

because the straight line is the shortest distance between two points, by Theorem 3.5. If we add these inequalities, we conclude that the inside approximation corresponding to a refinement is longer than the inside approximation corresponding to the original partition.

For the outside approximation corresponding to   A,   B,   and   C,   consider the tangents to the circle at   A,   B,   and   C.   Since we can assume that   A   and   B   are not diametrically opposed,   OA   and   OB   will be different lines that meet at   O,   so they will not be parallel, and they will have different slopes. As a result, since tangents are perpendicular to radii, by Theorem 3.7, the tangents at   A   and   C   whose slopes will be negative reciprocals of lines which have different slopes, so they will have different slopes, which is to say that they are not parallel, and they will meet at a point by Theorem 1.6. Call it   D, which by Theorem 5.1, will be outside the circle. Now consder the line which is tangent to the circle at   C.   Againt, it will meet both   AD   and   DB, outside the circle by Theorem 5.1 at points   E   and   F   respectively.

We will need to show that   E   is between   A   and   D,   and that   F   is between   D   and   F.

We will prove the following lemmas

For the rest of this discussion, we will assme that   H   is on the line segmen between   A   and   D.   (It could possibly be   D)   The other case could be handled similarly.

As a result, we can conclude that

|EF| < |ED| + |DF|

also since the straight line is the shortest distance between two points by Theorem 5.1, so

|AE| + |EF| + |FB| < |AE| + |ED| + |DF| + |FB| = |AD| + |DB|

which is to say that the outside approximation corresponding to   A,   B,   and   C   is shorter than the outside approximation corresponding just to   A   and   B.

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Next Theorem (5.3)

Analytic Foundations of Geometry

R. S. Wilson