B square of AC to the square of BC, that is, CD-AD2 : CEEB : : mo : n. Take a space S such that S : EB : : m : n. Then CD -AD + S: CE? :: m : n': that is, the difference between the square of CD and the given space AD-S, has to the square of CE a given ratio : and therefore, if m, n be unequal, the.locus of C is a given circle ;—if they be equal, it is a straight line. It is plain that if the circles be di. minished to points, this becomes the same as the third locus mentioned in the scholium, page 272. PROP. XIV. THEOR. If two straight lines in a given ratio, and containing a given angle, be drawn from a given point, and if one of them terminate in a given circumference, the other will terminate in another given circumference. Let A be a given point, and AB a straight line drawn from it to any point B in the circumference of a given circle: if another straight line be drawn through A, making with AB a given angle, and having to it a given ratio, its other extremity is in a given circumference. Join A and B with D, the centre of the given circle, and make the angle DAE equal to the given angle ; make AD to AE in the given ratio ; and make the angles EAC, AEC respectively equal to BAD, ADB: the circle of which E is the centre, and EC the radius, is the required locus. For since AD, which is given, has to AE a given ratio, AE is given : and (VI. 4.) AD : AB :: AE: AC; and, alternately, AD: AE:: AB: AC: therefore AB has to AC the given ratio. But the angles BAD, CAE being equal, add DAC to each; then BAC is equal to DAE, that is, (const.) to the given angle. In the same manner, it would be shown, that any other straight lines drawn through A, one to each circumference, and inclined at the given angle, would be in the given ratio. Therefore the circle of which E is the centre, and EC the radius, is the locus of C. PROP. XV. THEOR. Two concentric circles being given, if from any point A in the circumference of one of them, and from Draw AEF through the common centre, and to the radii AE, Join ED. Then (II. A.) AB? + AC + DB° + DC2 = 2AE2 + 2DE2 + 4EC4 = 2AE% + 2DE2 + 4AE.EG. But (II. 12.) ADo = AE + DE + 2AE.EG; wherefore ABP + AC2 + DBP + DC' = 2AD2: and the same may be shown to hold wherever D is taken in DG. Therefore DG is the required locus. D T PROP. XVI. PORIŞM. F K A Point may be found such that if, to any straight line whatever passing through it, perpendiculars be drawn from three given points, one of these perpendiculars is equal to the other two. Let A, B, C be three given points : join BC and bisect it in D; join AD, and take DE a third of it: if any straight line whatever FEGH be drawn through E, the perpendicular AG, drawn from one of the points A to FH, is equal to the two perpendiculars BF, CH, drawn from the points on the other side of FH. Draw DK perpendicular to FH, and through 5 D draw LDM parallel to FH, and meeting the perpendiculars LB, HC in L, M. Then, in the similar triangles AGE, DKE, AE: AG :: DE: DK; therefore, because AE is double of DE, AG is double of DK. Now, in the equiangular triangles DLB, DMC, BD, DC are equal, and therefore BL, CM are also equal. To each add CH : then LB, CH are equal to HM or KD. Add again FL; then BF, CH are equal to twice DK, that is, to AG. Schol. It is plain (APP. I. 7. schol.) that E is the point in which L EG H the straight lines drawn from the angles of the triangle ABC bisecting the opposite sides would cut one another. It may be farther remarked, that this porism may be regarded as the case in which the following problem becomes indeterminate : “Through a given point to draw a straight line, such that the perpendicular to it from one of three given points may be equal to the sum of those drawn to it from the other two;" '-a problem which is always determinate, except when the point through which the line is to be drawn, passes through E, as above determined ; since in every other case E is to be found in the manner already shown, and the required line is drawn through it and the given point. PROP. XVII. PORISM. A CIRCLE and a straight line being given, a point may be found within the circle, such that if any chord be drawn through it, the centre of the circle which passes through the extremities of the chord, and through the centre of the given circle, will be in the given straight line. Let ABC be a given circle, and DE a straight line given in position; a point F may be found, within the circle, such that the centre of a circle described through G, the given centre, and through the extremities of any chord AC, passing through F, will be in DE Draw GE perpendicular to DE, and take FG such that 2EG: BG :: BG: GF; F is the point through which the chords pass. Through F let any chord AC, be drawn, and draw GHD perpendicular to it, and join AG. Now (111. 31. and VI. 8. cor.) the square of AG is equal to the rectangle under GH, and twice the radius of a circle passing through A, G, C. But (const.) BG or AG® is equal to 2EG.GF; and (VI. 4.) DG : GE :: FG: GH; whence (VI. 16. and by doubling,) 2DG.GH is equal to 2EG.GF, and (I. ax. 1.) 2DG.GH is equal to AGʻ. Hence the rectangle under GH and twice DG is equal to the rectangle under GH and twice the radius of the circle passing through A, G, C: therefore DG must be the radius of that circle, and its centre is therefore in DE Schol. The property developed in this proposition, might also be expressed in form of the following local theorem: The locus of the centre of a circle described through the centre of a given circle, and the extremities of a chord passing through a given point, is a straight line given in position. G K G E PROP. XVIII. THEOR. Of isoperimetrical triangles on the same base, the isosceles one is the greatest. Let ABC be an isosceles triangle on the base BC; it is greater than any other triangle on BC, of equal perimeter, and not isosceles; that is, than any other triangle on BC, having the sum of its other sides equal to the sum of AB, AC. Draw DAE parallel and AF perpendicular to BC: the altitude of any triangle on BC, having its perimeter equal to that of ABC, and not being isosceles, is less than AF. For through any other point G in the parallel draw GB, GC. Then the angles which BA, CA make with DE are equal, being (I. 29.) equal to the equal angles ABC, ACB; while those which GB, GC make with it, are unequal, being equal to the unequal angles GBC, GCB. Hence (APP. 1. 12.) the sum of BG, GC is greater than the sum of BA, AC. Let GH be perpendicular to BC, and draw BK, CK to any point K in its continuation through G. Then (I. 19.) in the triangles BGK, CGK, BK is greater than BG, and CK than CG; wherefore BK, KC are greater than BG, GC, and consequently than BA, AC, which are less than BG, GC. Hence, by adding BC, the triangle BGC, which has its altitude equal to AF; and the triangle AKC, which has its altitude greater than AF, have each a greater perimeter than ABC: and therefore a triangle on BC, which is not isosceles, and which has its perimeter equal to that of ABC, must have its altitude less than AF, and must therefore be less than ABC.* сH PROP. XIX. THEOR. If two sides of a triangle be given, the triangle is a maximum, when they contain a right angle. Let ABC, ABD be triangles on the base AB, and having the sides BC, BD equal : if ABC be a right angle, the triangle ABC is greater than ABD. For (I. 19.) the perpendicular DE is less than DB, or than the perpendicular BC; and therefore, since the base AB is common, the triangle ABD is less than ABC. AE The demonstration of this proposition given in Legendre's Geometry, Book IV. Appendix, is deficient, as it is assumed without proof, that, in his diagram, the point M is not in the straight line AN. PROP. XX. THEOR. E Of isoperimetrical polygons, having a given number of sides, the maximum is the one which is equilateral. For, if possible, let ABCDE be the maximum polygon, and yet the side A E greater than AB. On BE describe the isosceles triangle BFE having the sum of its sides BF, FE equal to the sum of BA, AE. Then (APP. II. 18.) the triangle BFE is greater than BAE, and (. ax. 4.) the polygon FBCDE than ABCDE. Hence the latter is not a maximum, and the same may be shown in a similar manner, unless the sides be all equal. Hence one requisite in constituting a polygon, of a given perimeter and a given number of sides, a maximum, is that it must be equilateral. B PROP. XXI. THEOR. If all the sides of a quadrilateral or polygon except one be given in magnitude, the figure will be a maximum, when that remaining side is the diameter of a semicircle, and the figure is inscribed in that semicircle. Let ABCDE be the greatest polygon that can be contained by sides AB, BC, CD, DE given in magnitude, and AE not given : AE is the diameter of a semicircle, and the angular points of the polygon are on the arc of that semicircle. Join BE. Then, if ABCDE be not a semicircle, and therefore ABE not a right angle, by making it a right angle the triangle ABE (App. II. 19.) would be enlarged, as would also the whole polygon, the magnitude of the part BCDE not being changed by a change in the angle ABE. If the polygon be a maximum, therefore, ABE must be a right angle, and the segment ABE a semicircle. In the same manner it would be shown, by drawing lines from C and D to A and E, that ACE, ADE are right angles ; and the proof will be the same if there be six or any greater number of sides ; whence the truth of the proposition is manifest. Schol. There is but one semicircle which will contain the maximum polygon. For, suppose ABCDE to be a semicircle containing it; the angles at its centre subtended by the chords AB, B A |