Proof
Take a straight line and cut it at so that the rectangle on
, equals the square on (II.11, the golden section).
With centre and radius describe a circle; in it apply chord
equal to (IV.1). Join , . Because , is tangent to the circle through , ,
(III.37); by III.32 the tangent–chord angle equals the alternate
inscribed angle. Tracking the resulting angle relations (with I.5
for the isosceles base angles) gives the required ratio.
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Full neighborhood
Depends on (7)
- II.11Proposition II.11To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the…
- IV.1Proposition IV.1Into a given circle to fit a straight line equal to a given straight line which is not greater than the diameter of the…
- IV.5Proposition IV.5About a given triangle to circumscribe a circle.
- I.5Proposition I.5In isosceles triangles the angles at the base are equal to one another; and if the equal straight lines be produced…
- 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.32Proposition III.32If a straight line touch a circle, and from the point of contact there be drawn across, in the circle, a straight line…
- III.37Proposition III.37If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut…
Required by (dependents) (3)
- IV.11Proposition IV.11In a given circle to inscribe an equilateral and equiangular pentagon.
- XIII.8Proposition XIII.8If in an equilateral and equiangular pentagon straight lines subtend two adjacent angles, they cut one another in…
- XIII.9Proposition XIII.9If the side of the hexagon and that of the decagon inscribed in the same circle be added together, the whole straight…
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