Euclid Book 13 Proposition 9
Statement
If a side of a regular hexagon and that of a regular decagon inscribed in the same circle are added together, then the whole line segment has been cut in the golden ratio, and its longer segment is the side of the hexagon.
Computational Explanation
GeometricSceneA.,B.,C.,D.,E.,F.,G.,H.,I.,J.,K.,L.,M.,N.,P.,Q.,R.,{},CircleThrough[A.,B.,C.,D.,E.,F.,G.,H.,I.,J.,K.,L.,M.,N.,P.,Q.],GeometricAssertionStyle[Polygon[{A.,B.,C.,D.,E.,F.,G.,H.,I.,J.}],EdgeForm[{Green,Thickness[0.015`]}]],StylePolygon[K.,L.,M.,N.,P.,Q.],EdgeForm[{Pink,Thickness[0.015`]}],"Regular",Line[{B.,C.,R.}],EuclideanDistance[C.,R.]EuclideanDistance[K.,L.],
EuclideanDistance[B.,R.]
EuclideanDistance[C.,R.]
EuclideanDistance[C.,R.]
EuclideanDistance[C.,B.]
EuclideanDistance[B.,R.] EuclideanDistance[C.,R.] EuclideanDistance[C.,R.] EuclideanDistance[C.,B.] |
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Explanations
Let be a circle; of the figures inscribed in the circle let be the side of a decagon, that of a hexagon, and let them be in a straight line; I say that the whole straight line has been cut in extreme and mean ratio, and is its greater segment.
For let the centre of the circle, the point, be taken, let , , be joined, and let be carried through to .
Since is the side of an equilateral decagon, therefore the circumference is five times the circumference ; therefore the circumference is quadruple of .
But, as the circumference is to , so is the angle to the angle ;therefore the angle is quadruple of the angle .
And, since the angle is equal to the angle ,therefore the angle is double of the angle .
And, since the straight line is equal to , for each of them is equal to the side of the hexagon inscribed in the circle ,the angle is also equal to the angle ;therefore the angle is double of the angle .
But the angle was proved double of the angle ; therefore the angle is quadruple of the angle .
But the angle was also proved quadruple of the angle ; therefore the angle is equal to the angle .
But the angle is common to the two triangles and ; therefore the remaining angle is also equal to the remaining angle ;therefore the triangle is equiangular with the triangle .
Therefore, proportionally, as is to , so is to .
But is equal to . Therefore, as is to , so is to .
And is greater than ; therefore is also greater than .
Therefore the straight line has been cut in extreme and mean ratio, and is its greater segment.
ABC
ABC
BC
CD
BD
CD
For let the centre of the circle, the point
E
EB
EC
ED
BE
A
Since
BC
ACB
BC
AC
CB
But, as the circumference
AC
CB
AEC
CEB
AEC
CEB
And, since the angle
EBC
ECB
AEC
ECB
And, since the straight line
EC
CD
ABC
CED
CDE
ECB
EDC
But the angle
AEC
ECB
AEC
EDC
But the angle
AEC
BEC
EDC
BEC
But the angle
EBD
BEC
BED
BED
ECB
EBD
EBC
Therefore, proportionally, as
DB
BE
EB
BC
But
EB
CD
BD
DC
DC
CB
And
BD
DC
DC
CB
Therefore the straight line
BD
DC