第13卷命题 13 · 球内接正四棱锥的构造与性质

Let the diameter AB of the given sphere be set out, and let it be cut at the point C so that AC is double of CB; let the semicircle ADB be described on AB, let CD be drawn from the point C at right angles to AB, and let DA be joined; let the circle EFG which has its radius equal to DC be set out, let the equilateral triangle EFG be inscribed in the circle EFG, [IV. 2] let the centre of the circle, the point H, be taken, [III. 1] let EH, HF, HG be joined; from the point H let HK be set up at right angles to the plane of the circle EFG, [XI. 12] let HK equal to the straight line AC be cut off from HK, and let KE, KF, KG be joined. Now, since KH is at right angles to the plane of the circle EFG, therefore it will also make right angles with all the straight lines which meet it and are in the plane of the circle EFG. [XI. Def. 3] But each of the straight lines HE, HF, HG meets it: therefore HK is at right angles to each of the straight lines HE, HF, HG. And, since AC is equal to HK, and CD to HE, and they contain right angles, therefore the base DA is equal to the base KE. [I. 4] For the same reason each of the straight lines KF, KG is also equal to DA; therefore the three straight lines KE, KF, KG are equal to one another. And, since AC is double of CB, therefore AB is triple of BC. But, as AB is to BC, so is the square on AD to the square on DC, as will be proved afterwards. Therefore the square on AD is triple of the square on DC.

设给定球直径AB，在C点分割使AC=2CB。作半圆ADB，从C作CD垂直于AB，连接DA。

But the square on FE is also triple of the square on EH, [XIII. 12] and DC is equal to EH; therefore DA is also equal to EF. But DA was proved equal to each of the straight lines KE, KF, KG; therefore each of the straight lines EF, FG, GE is also equal to each of the straight lines KE, KF, KG; therefore the four triangles EFG, KEF, KFG, KEG are equilateral. Therefore a pyramid has been constructed out of four equilateral triangles, the triangle EFG being its base and the point K its vertex. It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. For let the straight line HL be produced in a straight line with KH, and let HL be made equal to CB. Now, since, as AC is to CD, so is CD to CB, [VI. 8, Por.] while AC is equal to KH, CD to HE, and CB to HL, therefore, as KH is to HE, so is EH to HL; therefore the rectangle KH, HL is equal to the square on EH. [VI. 17] And each of the angles KHE. EHL is right; therefore the semicircle described on KL will pass through E also.

作圆EFG半径等于DC，内接等边三角形EFG，取圆心H，连接EH、HF、HG。从H作HK垂直于圆平面，使HK=AC，连接KE、KF、KG。

[cf. VI. 8, III. 31.] If then, KL remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, since, if FL, LG be joined, the angles at F, G similarly become right angles; and the pyramid will be comprehended in the given sphere. For KL, the diameter of the sphere, is equal to the diameter AB of the given sphere, inasmuch as KH was made equal to AC, and HL to CB. I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid For, since AC is double of CB, therefore AB is triple of BC; and, convertendo, BA is one and a half times AC. But, as BA is to AC, so is the square on BA to the square on AD. Therefore the square on BA is also one and a half times the square on AD. And BA is the diameter of the given sphere, and AD is equal to the side of the pyramid. Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Q.

由于KH垂直于圆平面，故与HE、HF、HG成直角。由AC=HK、CD=HE及直角，得DA=KE。同理KF、KG等于DA，故KE、KF、KG相等。

E. D. LEMMA. It is to be proved that, as AB is to BC, so is the square on AD to the square on DC. For let the figure of the semicircle be set out, let DB be joined, let the square EC be described on AC, and let the parallelogram FB be completed. Since then, because the triangle DAB is equiangular with the triangle DAC, as BA is to AD, so is DA to AC, [VI. 8, VI. 4] therefore the rectangle BA, AC is equal to the square on AD. [VI. 17] And since, as AB is to BC, so is EB to BF, [VI. 1] and EB is the rectangle BA, AC, for EA is equal to AC, and BF is the rectangle AC, CB, therefore, as AB is to BC, so is the rectangle BA, AC to the rectangle AC, CB. And the rectangle BA, AC is equal to the square on AD, and the rectangle AC, CB to the square on DC, for the perpendicular DC is a mean proportional between the segments AC, CB of the base, because the angle ADB is right.

由AC=2CB得AB=3BC，且AB:BC等于AD²:DC²，故AD²=3DC²。又FE²=3EH²且DC=EH，故DA=EF。因此四个三角形均为等边，构成棱锥。