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in position, then is the triangle of least perimeter that whose sides AC, BC, are inclined to the line LM in equal angles.

For, let BM be drawn from B, perpendicularly to LM, and produced till DM BM: join AD, and from the point

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c where AD cuts LM draw BC: also, from any other point c', assumed in LM, draw c'a, c ́B, C'D. Then the triangles DMC, BMC, having the angle DCM = angle ACL (th. 7 Geom.) = мCв (by hyp.), DMC BMC, and DM BM, and MC common to both, have also DC BC (th. 1 Geom).

So also, we have c'D = C'B. Hence AC + CB = AC+CD AD, is less than AC' + c'D (theor. 10 Geom.), or than its equal Ac' + c'B. And consequently, AB + BC + AC is less than AB + BC' + AC'. Q. E. D.

Cor. 1. Of all triangles of the same base and the same altitude, or of all equal triangles of the same base, the isosceles triangle has the smallest perimeter.

For, the locus of the vertices of all triangles of the same altitude will be a right line LM parallel to the base; and when LM in the above figure becomes parallel to AB, since MCB = ACL, MCB CBA (th. 12 Geom.), ACL = CAB; it follows that CAB = CBA, and consequently AC = CB (th. 4 Geom.).

Cor. 2. Of all triangles of the same surface, that which has the minimum perimeter is equilateral.

For the triangle of the smallest perimeter, with the same surface, must be isosceles, whichever of the sides be considered as base: therefore, the triangle of smallest perimeter has each two or each pair of its sides equal, and consequently it is equilateral.

Cor. 3. Of all rectilinear figures, with a given magnitude and a given number of sides, that which has the smallest perimeter is equilateral.

For so long as any two adjacent sides are not equal, we may draw a diagonal to become a base to those two sides, and then draw an isosceles triangle equal to the triangle so cut off, but of less perimeter: whence the corollary is manifest.


To illustrate the second corollary above, the student may proceed thus: assuming an isosceles triangle whose base is not equal to either of the two sides, and then, taking for a new base one of those sides of that triangle, he may construct another isosceles triangle equal to it, but of a smaller perimeter. Afterwards, if the base and sides of this second isosceles triVOL. III.



angle are not respectively equal, he may construct a third isosceles triangle equal to it, but of a still smaller perimeter : and so on. In performing these successive operations, he will find that the new triangles will approach nearer and nearer to an equilateral triangle.


Of all Triangles of the Same Base, and of Equal Perimeters, the Isosceles Triangle has the Greatest Surface.

Let ABC, ABD, be two triangles of the same base AB and with equal perimeters, of which the one ABC is isosceles, the other is not: then the triangle ABC has a surface (or an altitude) greater than the surface (or than the altitude) of the triangle ABD.

Draw c'd through D, parallel to AB, to

cut CE (drawn perpendicular to AB) in c': then it is to be demonstrated that CE is greater than c'E.

The triangles AC'B, ADB, are equal both in base and altitude; but the triangle AC'B is isosceles, while ADB is scalene: therefore the triangle Ac'в has a smaller perimeter than the triangle ADB (th. 1 cor. 1), or than ACB (by hyp.). quently AC'<AC; and in the right-angled triangles AEC', AEC, having AE common, we have c'E < CE*. Q. E. D.

Cor. Of all isoperimetrical figures, of which the number of sides is given, that which is the greatest has all its sides equal. And in particular, of all isoperimetrical triangles, that whose surface is a maximum, is equilateral.

For, so long as any two adjacent sides are not equal, the surface may be augmented without increasing the perimeter.

Remark. Nearly as in this theorem may it be proved that, of all triangles of equal heights, and of which the sum of the two sides is equa., that which is isosceles has the greatest base. And, of all triangles standing on the same base and having equal vertical angles, the isosceles one is the greatest.

* When two mathematical quantities are separated by the character <, it denotes that the preceding quantity is less than the succeeding one: when, on the contrary, the separating character is >, it denotes that the preceding quantity is greater than the succeeding one.

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Of all Right Lines that can be drawn through a Given Point, between Two Right Lines Given in Position, that which is Bisected by the Given Point forms with the other two Lines the Least Triangle.


Of all right lines GD, AB, GD, can be drawn through a given point P to cut the right lines CA, CD, given in position, thrat, AB, which is bisected by the given point P, forms with CA, CD, the least triangle, ABC.





For, let EE be drawn through A parallel to CD, meeting DG (produced if necessary) in E; then the triangles PBD, PAE, are manifestly equiangular; and, since the corresponding sides PB, PA are equal, the triangles are equal also. Hence PBD will be less or greater than PAG, according as CG is greater or less than CA. In the former case, let PACD, which is common, be added to both; then will BAC be less than DGC (ax. 4 Geom.). In the latter case, if PGCB be added, pCG will be greater than BAC; and consequently in this case also BAC is less than DCG. Q. E. D.

Cor. If PM and PN be drawn parallel to CB and CA respectively, the two triangles PAM, PBN, will be equal, and these two taken together (since AMPNMC) will be equal to the parallelogram PMCN: and consequently the parallelo gram PMCN is equal to half ABC, but less than half DGC. From which it follows (consistently with both the algebraical and geometrical solution of prob. 8, Application of Algebra to Geometry), that a parallelogram is always less than half a triangle in which it is inscribed, except when the base of the one is half the base of the other, or the height of the former half the height of the latter; in which case the parallelogram is just half the triangle: this being the maximum parallelo gram inscribed in the triangle.


From the preceding corollary it might easily be shown, that the least triangle which can possibly be described about, and the greatest parallelogram which can be inscribed in, any curve concave to its axis, will be when the subtangent is equal to half the base of the triangle, or to the whole base of the parallelogram: and that the two figures will be in the ratio of 2 to 1. But this is foreign to the present enquiry.

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Of all Triangles in which two Sides are Given in Magnitude, the Greatest is that in which the two Given Sides are Perpendicular to each other.

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For, assuming for base one of the given sides, the surface is proportional to the perpendicular let fall upon that side from the opposite extremity of the other given side: therefore, the surface is the greatest when that perpendicular is the greatest; that is to say, when the other side is not inclined to that perpendicular, but coincides with it: hence the surface is a maximum when the two given sides are perpendicular to each other.

Otherwise. Since the surface of a triangle, in which two sides are given, is proportional to the sine of the angle included between those two sides; it follows, that the triangle is the greatest when that sine is the greatest: but the greatest sine is the sine total, or the sine of a quadrant; therefore the two sides given make a quadrantal angle, or are perpendicular to each other.

Q. E. D.


Of all Rectilinear Figures in which all the Sides except one are known, the Greatest is that which may be Inscribed in a Semicircle whose Diameter is that Unknown Side.

For, if you suppose the contrary to be the case, then whenever the figure made with the sides given, and the side unknown, is not inscribable in a semicircle of which this latter is the diameter, viz, whenever any one of the angles, formed by lines drawn from the extremities of the unknown side to one of the summits of the figure, is not a right angle; we may make a figure greater than it, in which that angle shall be right, and which shall only differ from it in that respect: therefore, whenever all the angles, formed by right lines drawn from the several vertices of the figure to the extremities of the unknown line, are not right angles, or do not fall in the circumference of a semicircle, the figure is not in its maximum state. Q. E. D.


Of all Figures made with Sides Given in Number and Magnitude, that which may be Inscribed in a Circle is the Greatest.



Let ABCDEFG be the polygon inscribed, and abcdefg a polygon with equal sides, but not inscribable in a circle; so that AB=ab, вc=bc, &c; it is affirmed that the polygon ABCDEFG


is greater than the polygon abcdefg.

Draw the diameter EP; join AP, PB; upon ab = AB make the triangle abp, equal in all respects to ABP; and join ep. Then, of the two figures edcbp, pagfe, one at least is not (by hyp.) inscribable in the semicircle of which ep is the diameter. Consequently, one at least of these two figures is smaller than the corresponding part of the figure APBCDEFG (th. 5). Therefore the figure APBCDEFG is greater than the figure apbcdefg: and if from these there be taken away the respective triangles APB, apb, which are equal by construction, there will remain (ax. 5 Geom.) the polygon ABCDEFG greater than the polygon abcdefg.

Q. E. D.


The Magnitude of the Greatest Polygon which can be contained under any number of Unequal Sides, does not at all depend on the Order in which those Lines are connected with each other.

For, since the polygon is a maximum under given sides, it is inscribable in a circle (th. 6). And this inscribed polygon is constituted of as many isosceles triangles as it has sides, those sides forming the bases of the respective triangles, the other sides of all the triangles being radii of the circle, and their common summit the centre of the circle. Consequently, -the magnitude of the polygon, that is, of the assemblage of these triangles, does not at all depend on their disposition, or arrangement around the common centre.


Q. E. D.

If a Polygon Inscribed in a Circle have all its Sides Equal, all its Angles are likewise Equal, or it is a Regular Polygon. For, if lines be drawn from the several angles of the polygon, to the centre of the circumscribing circle, they will divide the polygon into as many isosceles triangles as it has sides; and each of these isosceles triangles will be equal to either of the others in all respects, and of course they will


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