Proof
The argument mirrors III.7 with the point outside. Let be the
external point and the line through and the centre ,
meeting the circle at (near) and (far). For any other line
from meeting the circle at (near) and (far), I.20 gives
, and the SAS inequality I.24 again orders the
distances by the angles at . The "two lengths per secant"
ordering (concave/convex) follows by separating the near and far
intersections.
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Full neighborhood
Depends on (4)
- III.7Proposition III.7If on the diameter of a circle a point be taken which is not the centre, and from the point straight lines fall upon…
- I.20Proposition I.20In any triangle two sides taken together in any manner are greater than the remaining one.
- I.24Proposition I.24If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the…
- I.15Definition I.15A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among…
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