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.

top


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.

top


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.

top


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.

top


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.

top


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.

top