Return to Vignettes
of Ancient Mathematics
(general diagram)
(preliminary diagram: moving or still)
If a straight line is drawn through a circle and divides the circle
into unequal parts, and a segment
of a circle which is no larger than a semicircle (Is
this condition superfluous?) is stood perpendicular on
it, and the circular-arc of the segment standing on it is divided
into unequal parts, then the straight
line subtending the smaller circular arc is the smallest of all
the straight-lines falling from the same point to the larger circular-arc
of the initial circle.
The proof uses the Pythagorean theorem
repeatedly to reduces a 3D theorem to a planar projection and
prove several cases not mentioned in the statement of the theorem.
Case 1 (display)
(diagram
1) For let there be drawn a straight line BD in circle ABGD
and dividing the circle into unequal parts,
and let circular-arc BGD be larger than circular-arc BAD, and
(diagram 2) let
a segment of a circle BED which is no larger than a semicircle
be stood perpendicular on BD, and (diagram
3) let circular-arc BED be divided into unequal parts
at point E, and let circular-arc BE be smaller than circular-arc
ED, and let EB be joined. (diagram 4 = preliminary
diagram: moving or still)
I say that BE is smaller than all the straight-lines falling from
point E to circular-arc BGD.
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Case 1a (general
diagram or preliminary
moving diagram or still): BE < any line from E to arc BGD
(diagram
5) For let an altitude EZ be drawn from point E to the plane
of circle ABG. In fact, it will fall on the common section of
planes ABGD, BED, on straight-line BD, since segment BED is perpendicular
to circle ABGD. (diagram
6) And let the center of circle ABGD be taken and let it be
H, and let ZH be joined and extended to sides Q, K, and (diagram
7) let EL be extended from point E to circular-arc BGD, and
let ZL be joined.
And since EZ is perpendicular to the plane of circle ABGD,
it therefore makes right angles to all the straight-lines touching
it which are in the plane of circle ABGD. But each of ZB, ZL which
are in the plane of circle ABGD touches EZ. And, therefore, each
of angles BZE, LZE are right. And since BZ is less than ZL (this requires a fairly trivial generalization
of Euclid, Elements iii 7), therefore, the square of BZ
is less than the square of ZL. Let the square of EZ be added in
common. Therefore The squares of EZ, ZB are less than the squares
of EZ, ZL. But the square of BE is equal to the square of EZ,
ZB, while the square of LE is equal to the square of EZ, ZL. And
the square of BE is, therefore, less than the square of LE. Therefore,
BE is less than LE. We will similarly prove that BE is less than
all the straight-lines falling from point E to circular-arc BGD.
Therefore, BE is less than all the straight-lines falling from
point E to circular-arc BGD.
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Case 1b (general
diagram or preliminary
moving diagram or still): let L and G be points on BGD, with L nearer
to B than G, then EL < EG
(diagram 8) But
I say that of the straight-lines drawn from point A between points
B, K the line nearer to it is always less than the line further
away.
(diagram 9)
For let there be drawn some other line EG and let ZG be joined.
And since LZ is smaller than ZG, therefore the square of ZL
is smaller than the square of ZG. Let the square of ZE be added
in common. Therefore the squares of EZ, ZL are smaller than the
squares of EZ, ZG. But the squares of LZ, ZL are equal to the
square of LE, while the squares of GZ, ZE are equal to the square
of EG. And the square of LE is, therefore, smaller than the square
of EG. Therefore, LE is smaller than EG.
We will similarly prove that of the straight-lines drawn from
point A between points B, K, the line nearer to EB is always smaller
than the line further away.
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Case 1c (general
diagram or preliminary
moving diagram or still): EK (above the diameter QZHK) > all other
lines from E to arc BGD.
(diagram
10)And let EK, ED be joined. I say, in fact, that EK is the
largest of all the straight-lines falling from point E to circular-arc
KD, while ED is the smallest of all the straight lines drawn between
points D, K.
(diagram 11)For
since KZ is larger than GZ, therefore the square of KZ is larger
than the square of ZG. Let the square of ZE be added in common.
Therefore the squares of KZ, ZE, i.e. the square of EK, is larger
than the squares of EZ, ZG, i.e. the square of EG. Therefore KE
is larger than EG. We will similarly show, in fact, that EK is
larger than all the straight-lines falling from point E to circular-arc
KD. Therefore, of all the straight-lines falling from point E
to circular-arc KD, EK is the largest.
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Case 1d (general
diagram or preliminary
moving diagram or still): ED is the smallest line from E to a point
between K and D, and the lines from E to a point between D, K,
those to points nearer to D are smaller than those further.
(diagram
12)I say that ED is also the smaller than all the straight-lines
falling from point E between points K, D.
(diagram 13)For
let another line EM be drawn and let MZ be joined.
And since DZ is smaller than ZM, therefore the square of DZ
is also smaller than ZM. Let the square of ZE be added in common.
Therefore, the squares of EZ, ZK, i.e. the square of EDD, is smaller
than the squares of EZ, ZM, i.e. the square of EM. Therefore DE
is smaller than EM. We will similarly prove that ED is smaller
than all the straight-lines falling from point E between points
K, D, and that of the straight-lines drawn between points K, D,
the line nearer to it is smaller than the line further away.
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Case
2 general
diagram or preliminary
moving diagram or still): If BD is a diameter for any point G on the
circular-arc of circle ABGD other than B, D, BE < EG < ED.
(diagram
14) But let the line BD, drawn through circle ABGD, be a diameter,
and let the rest be supposed as the same. I say that EB is less
than all the straight-lines falling from point E to the circular-arc
of circle ABGD and that ED is the largest.
(diagram 15)
Given the same things constructed, since circular-arc DE is larger
than circular-arc EB, and altitude EZ is drawn, therefore DZ is
also larger to (sic) ZB. And BD is
the diameter of circle ABGD. Therefore the center of the circle
is on ZD. Therefore DZ is larger than ZG, but ZG is larger than
ZB. Thus the square of DZ is also larger than the square of ZG,
while the square of ZG is larger than the square of ZB. Let the
square of ZE be added in common. Therefore, the squares of DZ,
ZE, i.e. the square of ED, are larger than the squares of GZ,
ZE, i.e. the square of BE. Therefore, DE is larger than EG, and
EG is larger than EB. We will similarly prove that of all the
straight-lines falling from point E to the circular-arc of circle
ABGD, ED is the largest, while EB is smallest.
Therefore, ED is the largest of all the straight-lines falling
from point E to the circular-arc of circle ABGD, while EB is smallest.
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