Thabit Ibn Qurra, On Inscribing a Regular Heptagon in a Circle, props. 17, 18. Blue are added notes. Click on the diagram for the appropriate situation, or use the general diagram.

Theorem 17, dividing a line according to the needed specifications.

Theorem 18, the construction of the regular heptagon from the appropriately divided line.

For Abu Sahl's analysis and construction using conic sections
(10th cent.), cf. J.L. Berggren, __Episodes in the Mathematics
of Islam__ (New York, Springer, 1986).

(diagram
1)

So we postulate a square with A, B, G, D on it and we lead side
AB in a straight line in the direction of A to E and we join the
diagonal BG. And we fix the end point of the ruler at point D
and its other end point on line AE where it divides AE at point
Z and triangle ZAH is equal to triangle GTD. And from point T
we lead a line KTL parallel to AG.

Notes on the construction: this can be effected
through conic sections. Archimedes may have done that, but the
text as the Arabs (and we) have it treats the matter as a neusis
or convergence construction. One can see that the construction
is possible by considering that:

As H approaches A, ZAH is smaller than TDG

As H approaches G, ZAH is larger than TDG

(diagram
3)

Hence, if we wiggle DZ, Z eventually will hit a position so that ZAH = TDG.

(diagram
4)

Then I say that the area of AB by KB is equal to the square of
ZA, and that the area of ZK by AK is equal to the square of KB,
and that each of the two lines BK, ZA is longer than line AK.

AB*KB = AZ

^{2}

ZK*AK = KB^{2}

and again that each of the segments AZ, BK is greater than AK

(diagram
5)

And that is because the area of GD by TL is equal to the area
of ZA by AH.

GD*TL = ZA*AH since triangles EAZ
and TDG are equal and 1/2 GD*TL = 1/2 ZA*AH

Then the ratio of line GD, I mean AB, to ZA is as the ratio
of AH to TL.

GD : ZA = AH : TL and GD = AB =>

1. AB : ZA = AH : TL

(diagram
6)

And because each of the triangles ZAH, ZKT is similar to triangle
TLD,

(diagram 7)

the ratio of AH to TL is as the ratio of ZA to LD, I mean KB.

AH : TL = ZA : LD and LD = KB =>

2. AH : TL = ZA : KB

3. AB : ZA = ZA : KB (from (1) and (2)

(diagram
8)

And also the ratio of TL, I mean AK, to KT, I mean KB, is as the
ratio of LD, I mean KB, to ZK.

TL : KT = LD : ZK and TL = AK and KT = KB
(each since BG is the diagonal of a square, with TL = LG = AK)
and LD = KB =>

4 AK : KB = KB : ZK

(diagram
4)

Then the area of AB by KB is equal to the square of ZA, and the
area of ZK by AK is equal to the square of KB, and each of the
two lines ZA, KB is longer than line AK, and that is what we wanted.

(The last part is not actually proved, but
is obvious. We need a lemma: a*c = b^{2} and a < c
=> a < b < c. Hence, ZK > AK, so that by 4 and the
lemma, KB > AK; and AB > KB, so that by 3 and the lemma,
AZ > KB, and, as was just shown, KB > AK, whence AZ >
AK)

Lemma: a*c = b^{2} and a <
c => a < b < c.

proof: Let a*c = b^{2} and a < c.

Suppose b < a. Then b^{2} < a^{2}.

Let c-a = d, whence b^{2} = a*c = a^{2} + a*d
< a^{2}.

Suppose b > c. Then b^{2} > c^{2}.

Let c-a = d, whence b^{2} = a*c = c^{2} - c*d
> c^{2}

We now want construct a circle divided into seven equal parts.

(diagram
1)

Well let us draw AB with known end points and we distinguish on
it two points G, D where the area of AD by GD is equal to the
square of DB, and the area of GB by DB is equal to the square
of AD and each of lines AG, DB is longer than GD, as is known
in the previous construction.

(diagram
2)

And from lines AG, GD, DB, we construct a triangle GED with side
GE equal to line AG and side DE equal to line DB. And we join
AE, EB, and we circumscribe about triangle AEB circle AEBHZ, and
we lead out lines EG, ED in a straight line to the circumference,
let them fall on points Z, H. We join BZ and we lead out line
GT from G to the intersection.

(diagram
3)

Since sides AG, GE of triangle AGE are equal., angle EAG is equal
to angle AEG and arc AZ to arc EB.

(diagram
4)

And because the area of AD by GD is equal to the square of DB,
I mean DE, triangle AED is similar to triangle GED and angle DAE
is equal to angle GED and arc ZE is equal to arc EB

AD*GD = DB^{2} = DE^{2}
=> AD : DE = DE : GD and EDG is
common.

So, the three arcs EB, AZ, ZH are equal to each other.

(diagram
5)

And ZB is parallel to AE, and angle GAE, I mean GED, is equal
to angle DBT.

ZB is parallel to AE because arc BE = arc
AZ (or any of several other reasons). GAE
= DBT because of alternating interior
angles, and hence,

GED = DBT.

(diagram
6)

And because angle GED is equal to angle DBT and angle GDE to angle
TDB, and line ED to line DB, GD equals DT. and GE equals TB, and
one circle goes through the four points B, E, G, T.

The construction of the circle follows from
ED : DT = BD : DG, which follows from the homologous terms being
equal.

(diagram
7)

Then because the area of GB by DB is equal to the square of AG,
I mean EG, and line GB is equal to TE and DB is equal to DE, the
area of TE by ED is equal to the square of EG, triangle TEG is
similar to triangle GED and angle DGE is equal to angle ETG.

GB*DB = AG^{2} = EG^{2}
TE*DE

(since GB = TE or GD+DB = ED+DT, and DB = DE) =>

TE : EG = EG : ED and GED is common.

But angle DGE equals twice angle GAE, so that angle GTE equals
twice angle GAE.

DGE is the external
angle of GEA so that DGE
= GEA + GAE = 2GAE

(diagram
8)

And angle GTD is equal to angle DBE, so that angle DBE equals
twice angle GAE and arc AE is double arc EB.

GTD = DBE
because they are both angles of arc EG.

DBE = BED
because DB = DE.

And because angle DEB equals angle DBE, arc HB is also double arc EB.

(diagram
9)

And each of the arcs AE, BH is divided in two equal halves which
are equal and are equal to arc EB.

And so circle AEBHZ is divided into seven equal parts, and that is what we wanted.