Proof
Let be a diameter and drawn at right angles to at
(I.11). Suppose meets the circle at another point ; join . Since is a diameter and on the
circle, by III.31 (proved independently below) is
right. But is also right by construction; the sum of
two angles of is then two right angles, leaving no
positive angle at — contradicting I.17. Hence meets the
circle only at . The "no other line interposable" follows from
the uniqueness of the perpendicular (I.11): any line through
not perpendicular to makes a non-right angle and cuts the
circle.
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Full neighborhood
Depends on (5)
- I.11Proposition I.11To draw a straight line at right angles to a given straight line from a given point on it.
- I.17Proposition I.17In any triangle two angles taken together in any manner are less than two right angles.
- I.32Proposition I.32In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles,…
- III.31Proposition III.31In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less…
- III.2Definition III.2A straight line is said to touch a circle which, meeting the circle and being produced, does not cut the circle.
Required by (dependents) (8)
- III.17Proposition III.17From a given point to draw a straight line touching a given circle.
- IV.2Proposition IV.2In a given circle to inscribe a triangle equiangular with a given triangle.
- IV.3Proposition IV.3About a given circle to circumscribe a triangle equiangular with a given triangle.
- IV.4Proposition IV.4In a given triangle to inscribe a circle.
- IV.7Proposition IV.7About a given circle to circumscribe a square.
- IV.8Proposition IV.8In a given square to inscribe a circle.
- IV.12Proposition IV.12About a given circle to circumscribe an equilateral and equiangular pentagon.
- IV.13Proposition IV.13In a given pentagon, which is equilateral and equiangular, to inscribe a circle.
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