Euclid Book 13 Proposition 8
Statement
In a regular pentagon (ABCDE), two diagonals (, ) through different vertices cut each other in the golden ratio (, ), and their longer segments are equal to the side of the pentagon (EHEACH).
AC
EB
EB
EH
EH
HB
AC
CH
CH
HA
Computational Explanation
GeometricScene{{A.,B.,C.,D.,E.,H.},{}},{GeometricAssertion[{Polygon[{A.,B.,C.,D.,E.}]},"Regular"],Line[{{A.,H.,C.},{E.,H.,B.}}]},,,EuclideanDistance[E.,H.]EuclideanDistance[E.,A.]EuclideanDistance[C.,H.]
EuclideanDistance[E.,B.]
EuclideanDistance[E.,H.]
EuclideanDistance[E.,H.]
EuclideanDistance[H.,B.]
EuclideanDistance[A.,C.]
EuclideanDistance[C.,H.]
EuclideanDistance[C.,H.]
EuclideanDistance[H.,A.]
EuclideanDistance[E.,B.] EuclideanDistance[E.,H.] EuclideanDistance[E.,H.] EuclideanDistance[H.,B.] |
 |
Explanations
For in the equilateral and equiangular pentagon let the straight lines , , cutting one another at the point , subtend two angles taken in order, the angles at , ; I say that each of them has been cut in extreme and mean ratio at the point , and their greater segments are equal to the side of the pentagon.
For let the circle be circumscribed about the pentagon .
Then, since the two straight lines, are equal to the two , , and they contain equal angles, therefore the base is equal to the base , the triangle is equal to the triangle , and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.
Therefore the angle is equal to the angle ; therefore the angle is double of the angle .
But the angle is also double of the angle , inasmuch as the circumference is also double of the circumference ;therefore the angle is equal to the angle ; hence the straight line is also equal to , that is, to .
And, since the straight line is equal to , the angle is also equal to the angle .
But the angle was proved equal to the angle ; therefore the angle is also equal to the angle .
And the angle is common to the two triangles and ; therefore the remaining angle is 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 has been cut in extreme and mean ratio at , and the greater segment is equal to the side of the pentagon.
Similarly we can prove that has also been cut in extreme and mean ratio at , and its greater segment is equal to the side of the pentagon.
ABCDE
AC
BE
H
A
B
H
For let the circle
ABCDE
ABCDE
Then, since the two straight lines
EA
AB
AB
BC
BE
AC
ABE
ABC
Therefore the angle
BAC
ABE
AHE
BAH
But the angle
EAC
BAC
EDC
CB
HAE
AHE
HE
EA
AB
And, since the straight line
BA
AE
ABE
AEB
But the angle
ABE
BAH
BEA
BAH
And the angle
ABE
ABE
ABH
BAE
AHB
ABE
ABH
EB
BA
AB
BH
But
BA
EH
BE
EH
EH
HB
And
BE
EH
EH
HB
Therefore
BE
H
HE
Similarly we can prove that
AC
H
CH