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Draw BI making the angle CBI = BCD, and consequently IBD BDC; the two triangles IBD, IBC, will be isosceles, and we shall have IC - IB = ID. Hence the point I, the middle point of DC, is at equal distances from the three points B, C, D. For a similar reason, the point O, the middle of BA, is equally distant from the points A, B, C..

Now, suppose CA CA and the angle BCA


BCA; if

AB be joined, and the arcs A' C, A'B, produced till they meet in D', the arc D'CA' will be a semicircumference, as well as DCA; therefore, since we have CA' CA, we shall also have CD CD. But in the triangle CID', we have CI+ID' > CD' ; hence ID' > CD— CI, or ID' > ID.



In the isosceles triangle CIB, bisect the angle I by the arc EIF, which will also bisect BC at right angles. If a point L is assumed between I and E, the distance BL, equal to LC, will be less than BI; for it might be shown as in art. 41, that BL+LC<BI+IC; and, taking the half of each, that BL<BI. But in the triangle D'LC, we have D'L> D'C — CL, and still more D'L> DC — CI, or D'L> DI, or D'L>BI; consequently DL> BL. Hence, if in the arc EIF, we seek for a point equally distant from the three points B, C, D', it can be found only in the prolongation of EI towards F. Let I be the point required; we shall have D'I′ = BI′ = CI′; the triangles I' CB, I CD', IBD', being isosceles, we shall have the equal angles IBC ICB, IBD = ID'B, ICD'ID'C.


angles D'BC+ CBA' are equal to two right angles, and


are likewise equal to two right angles; therefore

D'BI+IBC + CBA' = 2,

BCI-ICD' + BCA' = 2.

Add together the two equations, observing that IBC



But the

BCI, and

D'BI-ICD' — BD'I—ID' C — CD'B= CA'B; and we shall 2FBC+ CлB+ CBA' + BCA' = 4.


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the triangle ABC (501) = 2-21BC; so that we have

area A'BC-2-2 angle IBC;

likewise, in the triangle ABC, we should have

area ABC= 2-2 angle IBC.

Now the angle IBC has already been proved to be greater than IBC; hence the area A'BC is less than ABC.

The same demonstration would lead to the same conclusion, Fig. 277. if, taking always the arc CA' = CA, the angle BCA' (fig. 277) were made less than BCA; hence ABC is the greatest of all those triangles, which have two sides given, and the third to be assumed at pleasure.

Fig. 278.

559. Scholium 1. The triangle ABC (fig. 278), the greatest of all those which have two given sides CA, CB, may be inscribed in a semicircle, the diameter of which is the chord of the third side AB; for O being the middle point of AB, the distances OC, OB, as we have seen, are equal; hence the circumference of a small circle, described from the point O as a pole, with the distance OB, will pass through the three points A, B, C. Moreover, the straight line AB is a diameter to this small circle; for the centre, which must be at once in the plane of the small circle, and (456) in the plane of the arc of the great circle BOA, must of necessity be found in the intersection of those two planes, which is the straight line BA; hence BA will be a diameter.

560. Scholium II. In the triangle ABC, the angle C being equal to the sum of the other two A and B, the sum of all the three angles must be double of the angle C. But (489) that sum is always greater than two right angles; hence C is always greater than one.

561. Scholium III. If the sides CB, CA, are produced till they meet in E, the triangle BAE will be equal to the fourth part of the surface of the sphere. For the angle



hence the three angles of the triangle BAE are equivalent to the four ABC, ABE, CAB, BAE, whose sum is equal to four right angles; therefore (505) the surface of the triangle

BAE =4-2 =

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which is the fourth part of the surface of the sphere.
562. Scholium IV. There could be no maximum, if the sum
of the two given sides CA, CB, were equal to, or greater than,
the semicircumference of a great circle. For, since the triangle
ABC must be capable of being inscribed in a semicircle of the
sphere, the sum of the two sides CA, CB, will be less (460) than
the semicircumference BCA, and consequently less than half the
circumference of a great circle.

The reason why there can be no maximum, when the sum of the two given sides is greater than the semicircumference of a

great circle, is that in this case the triangle continues to augment, as the angle contained by its two given sides augments; and at last, when this angle becomes equal to two right angles, the three sides are all in the same plane, and form a whole circumference; the spherical triangle has then increased to a hemisphere, but it has at the same time ceased to be a triangle.


563. Of all the spherical triangles, formed with a given side and a given perimeter, the greatest is that in which the two undetermined sides are equal.

Demonstration. Let AB (fig. 279) the given side be common Fig. 279. to the two triangles ACB, ADB, and let AC+ CB=AD+DB; we are to show that the isosceles triangle ACB, in which AC CB, is greater than ADB, which is not isosceles.



Since these triangles have the common part AOB, it will be sufficient to prove that the triangle BOD is less than AOC. Now, the angle CBA, equal to CAB, is greater than OAB; therefore (497) the side AO is greater than OB. Take OI OB, make OK = OD, and join KI; the triangle OKI (497) will be equal to DOB. Now, if the triangle DOB, or its equal KOI, is not admitted to be less than OAC, it must be either equal or greater; in both which cases, since the point I is between A and O, the point K must be found in the prolongation of OC, otherwise the triangle OKI would be contained in the triangle CAO, and therefore less than CAO. This granted, since the shortest way from C to A is CA, we have CK+KI + IA > CA. But

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CK OD-CO, AI = AO — OB, KI = BD; hence OD-CO + AO — OB + BD CA, or, by reduction, AD— CB+BD> CA, or AD+BD>CA + CB. But this inequality is at variance with the supposition of


hence the point K cannot fall in the prolongation of OC; consequently it falls between 0 and C, and the triangle KOI or its equal ODB is less than ACO; therefore the isosceles triangle ACB is greater than ADB having the same base and perimeter, which is not isosceles.

564. Scholium. The two last theorems are analogous to those of art. 63 and 69, of the appendix to section fourth; and from

them may be deduced, with regard to spherical polygons, the same consequences as we have obtained respecting plane polygons. The chief are as follows:

565. Among spherical polygons of the same perimeter and the same number of sides, that is the greatest which has its sides equal. The demonstration is the same as that of art. 301.

566. Among spherical polygons formed of known sides and one side taken at pleasure, the greatest is that which can be inscribed in a semicircle the diameter of which is equal to the chord of the undetermined side.

The demonstration is deduced from art. 559, in the manner exhibited in art. 303. It is requisite for the existence of a maximum, that the sum of the given sides be less than the semicircumference of a great circle.

567. Among spherical polygons formed of given sides, the greatest is that which can be inscribed in a circle of the sphere. The demonstration is the same as that of art. 303.

568. Among spherical polygons which have the same perimeter and the same number of sides, the greatest is that which has its angles equal, and its sides equal.

This results from the first and the third of the above propositions.


All the propositions relating to the maxima of spherical polygons, are also applicable to solid angles, of which these polygons are he measures.



Of the Regular Polyedrons.


569. There can be only five regular polyedrons.

Demonstration. For regular polyedrons were defined as having equal regular polygons for their faces, and all their solid angles equal. These conditions cannot be fulfilled except in a small number of cases.

1. If the faces are equilateral triangles, polyedrons may be formed of them, having solid angles contained by three of those triangles, by four, or by five: hence arise three regular bodies, the tetraedron, the octaedron, the icosaedron. No other can be formed with equilateral triangles; for six angles of such a triangle are equal to four right angles, and (356) cannot form a solid angle.

2. If the faces are squares, their angles may be arranged by threes: hence results the hexaedron or cube. Four angles of a square are equal to four right angles, and cannot form a solid angle.

3. In fine, if the faces are regular pentagons, their angles may likewise be arranged by threes; the regular dodecaedron will thus be formed.

We can proceed no farther; three angles of a regular hexagon are equal to four right angles; three of a heptagon are greater. Hence there can be only five regular polyedrons; three formed with equilateral triangles, one with squares, and one with pentagons.

570. Scholium. In the following problem, we shall show that these five polyedrons actually exist; and that all their dimensions may be determined, when one of their faces is known.


571. One of the faces of a regular polyedron, or only a side of it, being given, to construct the polyedron.

Solution. This problem admits of five cases, which we proceed to solve in succession.

Construction of the Tetraedron.

572. Let ABC (fig. 280) be the equilateral triangle which is Fig. 230. to form one of the faces of the tetraedron. At the point 0, the centre of this triangle, erect OS perpendicular to the plane ABC; let this perpendicular terminate in S, so that AS = AB; join SB, SC; the pyramid S-ABC will be the tetraedron required.

For, on account of the equal distances OA, OB, OC, the oblique lines SA, SB, SC, are equally removed from the perpendicular SO, and consequently equal to each other. One of them SA AB; hence the four faces of the pyramid S-ABC are tri

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