About a given circle to circumscribe an equilateral and equiangular pentagon
be the given circle; thus it is required to circumscribe an equilateral and equiangular pentagon about the circle ABCDE
be conceived to be the angular points of the inscribed pentagon, so that the circumferences AB
are equal; [IV. 11
] through A
be drawn touching the circle; [III. 16, Por.
] let the centre F
of the circle ABCDE
be taken [III. 1
], and let FB
Then, since the straight line KL
touches the circle ABCDE
, and FC
has been joined from the centre F
to the point of contact at C
, therefore FC is perpendicular to KL; [III. 18] therefore each of the angles at C is right.
For the same reason the angles at the points B, D are also right.
And, since the angle FCK
is right, therefore the square on FK
is equal to the squares on FC
For the same reason [I. 47
] the square on FK is also equal to the squares on FB, BK; so that the squares on FC, CK are equal to the squares on FB, BK, of which the square on FC is equal to the square on FB;
therefore the square on CK
which remains is equal to the square on BK
is equal to CK
And, since FB
is equal to FC
, and FK
common, the two sides BF, FK are equal to the two sides CF, FK; and the base BK equal to the base CK; therefore the angle BFK is equal to the angle KFC, [I. 8] and the angle BKF to the angle FKC.
Therefore the angle BFC
is double of the angle KFC
, and the angle BKC of the angle FKC.
For the same reason the angle CFD is also double of the angle CFL, and the angle DLC of the angle FLC.
Now, since the circumference BC
is equal to CD
, the angle BFC
is also equal to the angle CFD
. [III. 27
And the angle BFC
is double of the angle KFC
, and the angle DFC
of the angle LFC
; therefore the angle KFC is also equal to the angle LFC.
But the angle FCK
is also equal to the angle FCL
; therefore FKC
are two triangles having two angles equal to two angles and one side equal to one side, namely FC
which is common to them; therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; [I. 26
] therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC.
And, since KC
is equal to CL
, therefore KL
is double of KC
For the same reason it can be proved that HK is also double of BK.
is equal to KC
; therefore HK is also equal to KL.
Similarly each of the straight lines HG
can also be proved equal to each of the straight lines HK
; therefore the pentagon GHKLM is equilateral.
I say next that it is also equiangular.
For, since the angle FKC
is equal to the angle FLC
, and the angle HKL
was proved double of the angle FKC
, and the angle KLM double of the angle FLC, therefore the angle HKL is also equal to the angle KLM.
Similarly each of the angles KHG
can also be proved equal to each of the angles HKL
; therefore the five angles GHK
are equal to one another.
Therefore the pentagon GHKLM
And it was also proved equilateral; and it has been circumscribed about the circle ABCDE
. Q. E. F.