My Method for Squaring the Hyperbola is this:
L e t AB be one Asymptote of the Hyperbola EdC; and let AE and BC be parallel to th'other: Let also AE be to BC as 2 to 1; and let the Parallelogram ABDE equal 1. See Fig . 1. And note, that the Letter x every where stands for Multiplication.
Supposing the Reader knows, that EA. αζ. KH. βη. dθ. γχ. δλ. εμ. CB. &c. are in an Harmonic series, or a series reciproca primanorum sue arithmetice proportionalium (otherwise he is referr'd for satisfaction to the 87, 88, 89, 90, 91, 92, 93, 94, 95, prop. Arithm. Infinitor. Wallisij :)
I say
A
B
C
d
E
A
=
1
1
×
2
+
1
3
×
4
+
1
5
×
6
+
1
7
×
8
+
1
9
×
10
{\displaystyle ABCdEA={\frac {1}{1\times 2}}+{\frac {1}{3\times 4}}+{\frac {1}{5\times 6}}+{\frac {1}{7\times 8}}+{\frac {1}{9\times 10}}}
}
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \\\ \\\ \ \end{matrix}}\right\}\,}}
infinitum .
E
d
C
D
E
=
1
2
×
3
+
1
4
×
5
+
1
6
×
7
+
1
8
×
9
+
1
10
×
11
{\displaystyle EdCDE={\frac {1}{2\times 3}}+{\frac {1}{4\times 5}}+{\frac {1}{6\times 7}}+{\frac {1}{8\times 9}}+{\frac {1}{10\times 11}}}
E
d
C
y
E
=
1
2
×
3
×
4
+
1
4
×
5
×
6
+
1
6
×
7
×
8
+
1
8
×
9
×
10
{\displaystyle EdCyE={\frac {1}{2\times 3\times 4}}+{\frac {1}{4\times 5\times 6}}+{\frac {1}{6\times 7\times 8}}+{\frac {1}{8\times 9\times 10}}}
For (in Fig. 2, & 3) the Parallelog.
And (in Fig. 4.) the Triangl.
Note.
C
A
=
1
1
×
2
{\displaystyle CA={\frac {1}{1\times 2}}}
E
d
C
=
1
2
×
3
×
4
=
◻
d
D
−
◻
d
F
2
{\displaystyle EdC={\frac {1}{2\times 3\times 4}}={\frac {\Box dD-\Box dF}{2}}}
1
2
C
A
=
d
D
+
d
F
{\displaystyle {\tfrac {1}{2}}CA=dD+dF}
d
D
=
1
2
×
3
{\displaystyle dD={\frac {1}{2\times 3}}}
d
F
=
1
3
×
4
{\displaystyle dF={\frac {1}{3\times 4}}}
E
b
d
=
1
4
×
5
×
6
=
◻
b
r
−
◻
b
n
2
{\displaystyle Ebd={\frac {1}{4\times 5\times 6}}={\frac {\Box br-\Box bn}{2}}}
1
2
d
D
=
b
r
+
b
n
{\displaystyle {\tfrac {1}{2}}dD=br+bn}
b
r
=
1
4
×
5
{\displaystyle br={\frac {1}{4\times 5}}}
b
n
=
1
5
×
6
{\displaystyle bn={\frac {1}{5\times 6}}}
d
f
C
=
1
6
×
7
×
8
=
◻
f
G
−
◻
f
k
2
{\displaystyle dfC={\frac {1}{6\times 7\times 8}}={\frac {\Box fG-\Box fk}{2}}}
1
2
d
F
=
f
G
+
f
k
{\displaystyle {\tfrac {1}{2}}dF=fG+fk}
f
G
=
1
6
×
7
{\displaystyle fG={\frac {1}{6\times 7}}}
f
k
=
1
7
×
8
{\displaystyle fk={\frac {1}{7\times 8}}}
E
a
b
=
1
8
×
9
×
10
=
◻
a
q
−
◻
a
p
2
{\displaystyle Eab={\frac {1}{8\times 9\times 10}}={\frac {\Box aq-\Box ap}{2}}}
1
2
b
r
=
a
q
+
a
p
{\displaystyle {\tfrac {1}{2}}br=aq+ap}
a
q
=
1
8
×
9
{\displaystyle aq={\frac {1}{8\times 9}}}
a
p
=
1
9
×
10
{\displaystyle ap={\frac {1}{9\times 10}}}
b
c
d
=
1
10
×
11
×
12
=
◻
c
s
−
◻
c
m
2
{\displaystyle bcd={\frac {1}{10\times 11\times 12}}={\frac {\Box cs-\Box cm}{2}}}
1
2
b
n
=
c
s
+
c
m
{\displaystyle {\tfrac {1}{2}}bn=cs+cm}
c
s
=
1
10
×
11
{\displaystyle cs={\frac {1}{10\times 11}}}
c
m
=
1
11
×
12
{\displaystyle cm={\frac {1}{11\times 12}}}
d
e
f
=
1
12
×
13
×
14
=
◻
e
t
−
◻
e
l
2
{\displaystyle def={\frac {1}{12\times 13\times 14}}={\frac {\Box et-\Box el}{2}}}
1
2
f
G
=
e
t
+
e
l
{\displaystyle {\tfrac {1}{2}}fG=et+el}
e
t
=
1
12
×
13
{\displaystyle et={\frac {1}{12\times 13}}}
e
l
=
1
13
×
14
{\displaystyle el={\frac {1}{13\times 14}}}
f
g
C
=
1
14
×
15
×
16
=
◻
g
u
−
◻
g
h
2
{\displaystyle fgC={\frac {1}{14\times 15\times 16}}={\frac {\Box gu-\Box gh}{2}}}
1
2
f
k
=
g
u
+
g
h
{\displaystyle {\tfrac {1}{2}}fk=gu+gh}
g
u
=
1
14
×
15
{\displaystyle gu={\frac {1}{14\times 15}}}
g
h
=
1
15
×
16
{\displaystyle gh={\frac {1}{15\times 16}}}
&c.
&c.
&c.
&c.
And
