Proof
Let touch the circle at , and be a chord from into
the circle. We show equals the inscribed angle in
the alternate segment .
Draw the diameter from . By III.18, the tangent is
perpendicular to , so right. By III.31, the
inscribed angle in the semicircle is right. In
, by I.32, two right angles; since is right, one right angle.
Now right (from the identity above). By III.21,
equals any other inscribed angle in the same segment as .
Hence equals the inscribed angle in the alternate
segment. The other angle on the tangent's other side
equals the inscribed angle in the original segment by analogous
argument.
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Full neighborhood
Depends on (6)
- III.18Proposition III.18If a straight line touch a circle, and a straight line be joined from the centre to the point of contact, the straight…
- III.20Proposition III.20In a circle the angle at the centre is double of the angle at the circumference, when the angles have the same…
- III.21Proposition III.21In a circle the angles in the same segment are equal to one another.
- 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…
- 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,…
- 3Common notion 3If equals be subtracted from equals, the remainders are equal.
Required by (dependents) (4)
- III.33Proposition III.33On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilineal angle.
- III.34Proposition III.34From a given circle to cut off a segment admitting an angle equal to a given rectilineal angle.
- IV.2Proposition IV.2In a given circle to inscribe a triangle equiangular with a given triangle.
- IV.10Proposition IV.10To construct an isosceles triangle having each of the angles at the base double of the remaining one.
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