Euclid Book 13 Proposition 10
Statement
If a regular pentagon is inscribed in a circle, then the square of the side of the pentagon equals the sum of the squares on the sides of a regular hexagon and a regular decagon inscribed in the same circle.
Computational Explanation
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2
EuclideanDistance[R.,S.]
2
EuclideanDistance[K.,L.]
2
EuclideanDistance[A.,B.]
2 EuclideanDistance[R.,S.] 2 EuclideanDistance[K.,L.] 2 EuclideanDistance[A.,B.] |
 |
Explanations
Let be a circle, and let the equilateral pentagon be inscribed in the circle .
I say that the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the circle .
For let the centre of the circle, the point, be taken, let be joined and carried through to the point , let be joined, let be drawn from perpendicular to and be carried through to , let , be joined, let be again drawn from perpendicular to , and be carried through to , and let be joined.
Since the circumference is equal to the circumference , and in them is equal to , therefore the remainder, the circumference , is equal to the remainder . But belongs to a pentagon; therefore belongs to a decagon.
And, since is equal to , and is perpendicular, therefore the angle is also equal to the angle .
Hence the circumference is also equal to ;therefore the circumference is double of the circumference ; therefore the straight line is a side of a decagon.
For the same reason is also double of . Now, since the circumference is double of the circumference , while the circumference is equal to the circumference , therefore the circumference is also double of the circumference .
But the circumference is also double of ; therefore the circumference is equal to the circumference . But is double of , since is so also; therefore is also double of . But, further, the circumference is also double of the circumference , for the circumference is equal to .
Therefore the whole circumference is also double of ; hence the angle is also double of the angle .
But the angle is also double of the angle , for the angle is equal to the angle .
Therefore the angle is also equal to the angle .
But 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 the straight line is to , so is to ;therefore the rectangle , is equal to the square on .
Again, since is equal to , while is common and at right angles, therefore the base is equal to the base ;therefore the angle is also equal to the angle .
But the angle is equal to the angle ; therefore the angle is also equal to the angle .
And the angle at 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 the straight line is to , so is to ;therefore the rectangle , is equal to the square on .
But the rectangle, was also proved equal to the square on ; therefore the rectangle , together with the rectangle , , that is, the square on , is equal to the square on together with the square on .
And is a side of the pentagon, of the hexagon , and of the decagon.
ABCDE
ABCDE
ABCDE
I say that the square on the side of the pentagon
ABCDE
ABCDE
For let the centre of the circle, the point
F
AF
G
FB
FH
F
AB
K
AK
KB
FL
F
AK
M
KN
Since the circumference
ABCG
AEDG
ABC
AED
CG
GD
CD
CG
And, since
FA
FB
FH
AFK
KFB
Hence the circumference
AK
KB
AB
BK
AK
For the same reason
AK
KM
AB
BK
CD
AB
CD
BK
But the circumference
CD
CG
CG
BK
BK
KM
KA
CG
KM
CB
BK
CB
BA
Therefore the whole circumference
GB
BM
GFB
BFM
But the angle
GFB
FAB
FAB
ABF
Therefore the angle
BFN
FAB
But the angle
ABF
ABF
BFN
AFB
BNF
ABF
BFN
AB
BF
FB
BN
AB
BN
BF
Again, since
AL
LK
LN
KN
AN
LKN
LAN
But the angle
LAN
KBN
LKN
KBN
And the angle at
A
AKB
AKN
Therefore the remaining angle
AKB
KNA
KBA
KNA
Therefore, proportionally, as the straight line
BA
AK
KA
AN
BA
AN
AK
But the rectangle
AB
BN
BF
AB
BN
BA
AN
BA
BF
AK
And
BA
BF
AK