**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 *T*_{1},
*T*_{2}, . . . *T*_{n}, so we can take the outside
approximation to be

*|AT*_{1}| + |*T*_{1}*P*_{1}| + |*P*_{1}*T*_{2}| + |*T*_{2}*P*_{2}| + . . . + |*T _{n}B*|

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 *P*_{i }to *P*_{i+1} will be shorter than the distance going through *T*_{i+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

**Lemma 1**:*OC*intersects*AB*at a point*G*which is on the line segment between*A*and*B*.**Lemma 2**: There is a point*H*on the line determined by*OC*which is either on the line segment between*A*and*D*, on the line segment between*D*and*B*, or is equal to point*D*.

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.

**Lemma 3**:*E*is on the line segment between*A*and*H*.**Lemma 4**:*E*is on the line segment between*A*and*D*.**Lemma 5**:*F*is on the line segment between*D*and*B*.**Lemma 6**:*C*is on the line segment between*E*and*F*.

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*.

Analytic Foundations of Geometry