In the construction of the figure it is necessary to show, 1. that CGB = / ADB, 2. that CGB = CBG, 3. that. BC = CG, 4. that CB, BI, IG, GC = each other, and .. c is a 5. that □ c1 is rectangular, (square, i. e., CB3, 6. that similarly HF is a square, i. e., ac2. The Demonstration itself consists in showing that Ab2, i. e., fig. AE = ac2 + CB2 + figs. Ag + Ge, and that these are = 4c2 + CB3 + 2 AC × CB. PROPOSITION V. Theorem. If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle A contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line. D L H M K E F G Proved by showing that BC2 = figs. CM + HG +LF; and that these = figs. AH + LF; and these = AD X DB + CD2. part of it produced, together with the square of half the line bisected, is equal to the square of the line made up of the half and the part produced. Proved by showing that co2 = figs. LG; and these and that these last CHEF + AL + CM + LG, which AM + LG, PROPOSITION VII. Theorem. If a straight line be di vided into any two parts, the squares of the whole line and of one of the parts are equal to twice the rectangle contained by the whole line and that part, together with the square of the other part. Proved by showing that AB + BC2 = figs. AG + GE +2 CH + KF, and that these 2 AH + KF; and that these 2 AB × BC + AC2. PROPOSITION VIII. Theorem. If a straight line be divided into any two parts, four times the rectangle con tained by the whole line and one of the parts, together with the square of the other part, is equal to the square of the line made up of the whole and that part. In the construction of the figure it is necessary to show that BD = GK, in order to prove that the square BN Square GR, and also to show that BK KR, in order to prove that " AKOTM MR. -2 The Demonstration itself consists in showing that AB + BC2, i. e., the fig. AF = figs. AK + MP + PI + KF + BN + GR + XH, and that these are 2 AK + 2 MP + 2 GR + XH; and that these 4 AK + XH, and that these last 4 AB X BC + AC2. PROPOSITION IX. Theorem. If a straight line be divided into two In the construction of the figure it is necessary to prove, 1. that AEB is a right, which is done by showing that each of the S CEB, AEC = 1⁄2 rt. 2, FG, by showing that S GEF, Efg, eachrt. 4, and . ea. other. 3. that DF = DB, by showing that B and 2. that GE each = rt. 4, BFD andea. other. The Demonstration itself consists in showing that AD2 + DB2 = AF2; and that this = Ac2 + CE2 + EG2 + GF2, and that these are 2 ac2 + 2 cd2. PROPOSITION X. Theorem. If a straight line be bisected, and produced to any point, the square of the whole line thus. produced, and the square of duced. up E B In the construction of the figure it is necessary to show, کردے 2. that AEB is a rt. 4, 1. that S BEF + EFD < 2 rt. Zs, in order to prove that EB, FD will meet, if produced, as in G; also by showing that each of the S CEA, CEB = right, 3. that BD DG, by showing that S DBG, DGB each right, and.. = each other. 4. that GFFE, by showing that GEF = Fge. = The Demonstration itself consists in showing that AD2 + DB2 = AG'; and that this Ac2 + CE2 + EF2 + FG, and that these =2 AC2 + 2 CD2. PROPOSITION XI. Problem. To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square of the other part. Steps of the Demonstration. 1. Prove that CF X FA + AE2 EF2, and that this 2. = AE2 + AB2, 3. that CF X FA, i. e., fig FK = AB2, i. e., fig. ad, that FH HD, 4. 3. that. AB X BH = AH2. PROPOSITION XII. Theorem. In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle, between the perpendicular and the obtuse angle. B Obs. This proposition is so short, and each step so necessary to the proof, that to attempt to simplify it would be useless. The learner will find no difficulty in it whatever, if he can remember that BC2 + CD + 2 BC X CD is to be substituted for DB2; but this should be carefully noticed, as it is very likely to escape the memory. PROPOSITION XIII. Theorem. In every triangle, the square of the side subtending any of the acute angles is less than the |