Proof
Let be the given circle. Draw any chord in it (Postulate
1) and bisect at (I.10). From draw at right
angles to (I.11), produced to meet the circle at and .
Bisect at (I.10); then is the centre. For if any other
point were the centre, then by SSS (I.8) on and
we would obtain , both
right (I.13). But already lies on the perpendicular bisector
of , and the perpendicular at is unique (I.11); applying
the same reasoning to chord forces onto its perpendicular
bisector as well. The two perpendicular bisectors meet only at the
true centre, which is .
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Full neighborhood
Depends on (6)
- I.8Proposition I.8If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they…
- I.10Proposition I.10To bisect a given finite straight line.
- I.11Proposition I.11To draw a straight line at right angles to a given straight line from a given point on it.
- I.13Proposition I.13If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two…
- 1Postulate 1To draw a straight line from any point to any point.
- 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…
Required by (dependents) (4)
- III.2Proposition III.2If on the circumference of a circle two points be taken at random, the straight line joining the points will fall…
- III.3Proposition III.3If in a circle a straight line through the centre bisect a straight line not through the centre, it also cuts it at…
- III.9Proposition III.9If a point be taken within a circle, and more than two equal straight lines fall from the point on the circle, the…
- III.17Proposition III.17From a given point to draw a straight line touching a given circle.
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