Gen. Enun.-If a perpendicular be drawn bisecting a given straight line, any point in this perpendicular is at equal distances, and any point without the perpendicular is at unequal distances from the extremities of the line. Part. Enun.--Let AB be a given st. line. Bisect it in c, and through c draw cd at rt. Ls to it. Then any pt. d upon the 1 CD js =ly distant from A and B; and any pt. E, without the I, is not = ly distant from A and B. CONST.-Join AD, BD. DEMONST.-1. Because AC = CB, and cd is common to AS ACD, BCD ; .. the sides AC, CD = BC, CD, each to each, and the rt. L rt. BCD; .. the base AD the base BD (Prop. IV.): and the same may be proved of the lines drawn from any other pt. in co to A and B. 2. Again, if E be a pt. without the 1 , join EA, EB, DB. Then, as before, AD = DB; .. AE = AD + DE = BD + DE, But BD + DE are > BE (Prop. XX.); .. AE is > BE: and the same may be proved of any other pt. on either side of cd. Wherefore, if a I be drawn &c.-Q. E. D. C B в ACD PROP. P. PROB. Gen. Enun.-From two given points on the same side of a line given in position, to draw two lines which shall meet in a point in this line, so that their sum shall be less B F E any other than that of any other lines drawn from the two given points to any other point in the same line. PART. ENUN.—Let all be a line gn. in position, and c, D two gn. pts. on the same side of it. Then it is required to draw from the pts. C, D, two st. lines which shall meet in a pt. in AB, and of which the sum shall be less than that two lines drawn from c and D to any other pt. in AB. Const.-From c and draw ce, DF, I to AB (Prop. XII.); produce ce to g, making EG = CE (Prop. 111.); join dg, cutting AB in H, and join ch. Then ch, Du are 'the lines required. In ab take any other pt. K, and join CK, DK, GK. DemonsT.-1. Now in the ASCEH, GEH, the two sides CE, EH = GE, EH, each to each, and the rt. L cez = rt. 2 GEH (Ax. 10); .. the base ch = base Gh. (Prop. IV.) 2. In the same way it may be proved that ck = GK. 3. Again, because ch = GH, add to each dh; .. CH + DH = GH + DH (Ax. 2) = GD; but gd is < DK + KG (Prop. XX.), and Kg = ck; .. CH + ph is < CK + DK. 4. In like manner it may be shown that ch + dh is < the sum of any other two lines drawn from c and to any pt. in AB. Wherefore, from two gn. pts. C, D, &c.-Q. E. F. PROP. XXI. THEOR. GEN. ENUN.-If, from the ends of one side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. PART. ENUN.-Let ABC be a A. From B, C, the extremities of the side wc, draw the st. lines BD, CD, to the pt. D within the a. Then BD + A. B с CD are < BA + AC; but the 2 BDC is > the Z BAC. DEMONST.–1. Produce bo to E; and because two sides of a A are > the third side, .. in A ABE, BA + AE are > BE. (Prop. XX.) Add to each of these EC; . . BA + AC are > BE + EC. 2. Also in the A CED, DE + Ec are > DC. Add to each of these BD; . . BE + EC are >than BD + DC; and .., a fortiori, ba + Ac are > BD + DC. 3. Again, because the exterior < of a A is > the interior opposite < (Prop. XVI.), ... the exterior 2 BDC of A CDE is > interior opposite Z CED; and the exterior L CED of A ABE is > interior opposite Z BAE ; •., a fortiori, 2 BDC is > the 2 BAE. Wherefore, if from the ends &c.-Q. E. D. It is absolutely essential to the truth of this Theorem that the two lines should be drawn from the extremities of one of the sides; for the mere fact that one of the triangles is included within the other is not sufficient. The Proposition is indeed by no means self-evident, as some have supposed ; for it will appear from the two subjoined problems that in certain cases the sum of two lines drawn within a A may either exceed that of the two sides of the A, or contain a smaller L. PROP. Q. PROB. Gen. Enun.-In a right-angled or obtuse-angled triangle to determine a point, to which two straight lines being drawn, one from the extremity of the base farthest from the right angle or obtuse angle, and the other from any point upon the base, their sum shall exceed that of the two sides of the triangle. Α. PART. ENUN.—Let ABC be a A, having the at B either a rt. L or an obtuse L. Then it is required to find a pt. within it, to which two lines being drawn from the extremity c, and any pt. D, upon the base BC, their sum shall exceed that of AB + AC. Const.-Join AD. Because any two 28 of a A are together less than two rt. 2 * (Prop. XVII.); and that the 2 ABD is either a rt. or > a rt. L (Hyp.); .. the L ADB is < a rt. L, and consequently < the 4 ABD ; ..B Ad is > than AB. (Prop. XIX.) Make = AB (Prop. III.), and bisect AE in F. (Prop. X.) Then F is the pt. required. Join Fc. DEMONST.-1. Because AF=FE,.. AF + FC=EF + FC; but AF + FC are > AC (Prop. XX.); .. EF + Fc are > AC. 2. Now ED = AB; .. by addition, DF + FC are > AB + AC (Ax. 4). Therefore in the A ABC, a pt. F has been found such that DF + fc is > AB + AC.-Q. E. F. с ED B D PROP. R. PROB. A Gen. Enun.-From one side of a scalene triangle, to draw two straight lines to a point within the triangle, which shall contain an angle less than that which is contained by the other two sides of the triangle. PART. ENUN.—Let ABC be a scalene A, of which al is the shortest and bc the longest side. Then it is required, &c. CONST.-From Bc cut off DB = AB. (Prop. III.) Join AD: and in AD take any pt. E, and join Ec. Then the L CED shall be < than L BAC. DEMONST.– Since BA = BD, .. the BAD = BDA (Prop. V.); but the exterior, BDA is > the interior opposite / CED (Prop. XVI.); :: also, the / BAD is > the / CED; and ..., a fortiori, / BAC is > the L ced. B с D Therefore from the base Bc of the A ABC, two lines ce, DE have been drawn to the pt. E, containing the CED < the L BAC.-Q. E. F. From Prop. XXI. the following is a deduction: PROP. S. THEOR. E D Gen. Enun.-If a trapezium and a triangle are upon the same base, and on the same side of it, and the trapezium falls within the triangle, the sides of the trapezium are together less than those of the triangle. PART. ENUN.- Let the A ABC and the trapezium ABED be upon the same base AB, and upon the same side of it, and let the trapezium fall within the A. Then the sides AD + DE + EB are < AC + CB. DEMONST.-Produce AD, BE to meet in F. Then AF + BF are < AC + BC (Prop. XXI.); and DF + EF are > than DE (Prop. XX.); .., a fortiori, AD + DE + EB are < than AC + BC. Wherefore, if a trapezium &c.-Q. E. D. B PROP. XXII. PROB. Gen. ENUN.—To make a triangle of which the sides shall be equal to three given straight lines; but any two whatever of these lines must be greater than the third. Part. Enun.—Let A, B, c be the gn. st. lines, of which any two are > the third ; viz. A + B > C, A + c > B, and B + c > Then it is required to make a A of which the A. |