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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


have the angles at their bases all equal: consequently, the angles of the polygon, which are each made up of two angles at the bases of two contiguous isosceles triangles, will be equal to one another. Q. E. D.


Of all Figures having the Same Number of Sides and Equal Perimeters, the Greatest is Regular.

For, the greatest figure under the given conditions has all its sides equal (th. 2 cor.). But since the sum of the sides and the number of them are given, each of them is given therefore (th. 6), the figure is inscribable in a circle: and consequently (th. 8) all its angles are equal; that is, it is regular. Q. E. D.

Cor. Hence we see that regular polygons possess the property of a maximum of surface, when compared with any other figures of the same name and with equal perimeters.


A Regular Polygon has a Smaller Perimeter than an Irregular one Equal to it in Surface, and having the Same Number of Sides.

This is the converse of the preceding theorem, and may be demonstrated thus: Let R and I be two figures equal in surface and having the same number of sides, of which R is regular, I irregular: let also R be a regular figure similar to R, and having a perimeter equal to that of 1. Then (th. 9) R>1; but I = R; therefore R' > R. But R' and R are similar; consequently, perimeter of R'> perimeter of R; while R' per. I (by hyp.). Hence, per. I > per. R. Q. E. D. per.



The Surfaces of Polygons, Circumscribed about the Same or Equal Circles, are respectively as their Perimeters*. Let the polygon ABCD be circumscribed about the circle EFGH; and let this polygon be divided into triangles, by lines drawn. from its several angles to the centre o of the circle. Then, since each of the tangents AB, BC, &c, is perpendicular to its



*This theorem, together with the analogous ones respecting bodies circumscribin cylinders and spheres, were given by Emerson in his Geometry, and their use in the theory of Isoperimeters was just suggested: but the full application of them to that theory is due to Simon Lhuillier.


corresponding radius OE, OF, &c, drawn to the point of contact (th. 46 Geom.); and since the area of a triangle is equal to the rectangle of the perpendicular and half the base (Mens. of Surfaces, pr. 2); it follows, that the area of each of the triangles ABO, BCO, &c, is equal to the rectangle of the radius of the circle and half the corresponding side AB, BC, &c: and consequently, the area of the polygon ABCD, circumscribing the circle, will be equal to the rectangle of the radius of the circle and half the perimeter of the polygon. But, the surface of the circle is equal to the rectangle of the radius and half the circumference (th. 94 Geom.). Therefore, the surface of the circle, is to that of the polygon, as half the circumference of the former, to half the perimeter of the latter; or, as the circumference of the former, to the perimeter of the latter. Now, let P and P' be any two polygons circumscribing a circle c: then, by the foregoing, we have

surf. c surf. P:: circum. c: perim. P,


surf. c: surf. p' :: circum. c: perim. P.

But, since the antecedents of the ratios in both these proportions, are equal, the consequents are proportional: that is, surf. P: surf. P':: perim. P: perim. P'. Q. E. D.

Cor. 1. Any one of the triangular portions ABO, of a polygon circumscribing a circle, is to the corresponding circular sector, as the side AB of the polygon, to the arc of the circle included between AO and BO.

Cor. 2. Every circular arc is greater than its chord, and less than the sum of the two tangents drawn from its extremities and produced till they meet.

The first part of this corollary is evident, because a right line is the shortest distance between two given points. The second part follows at once from this proposition: for EA + AH being to the arch EIH, as the quadrangle AEOH to the circular sector HIEO; and the quadrangle being greater than the sector, because it contains it; it follows that EA+AH is greater than the arch EIH*.

Cor. 3. Hence also, any single tangent EA, is greater than its corresponding arc EI.

This second corollary is introduced, not because of its immediate connection with the subject under discussion, but because, notwithstanding its simplicity, some authors have employed whole pages in attempting its de monstration, and failed at last.


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