Construction.-From A draw AE at right angles to AB, and from AE cut off AF equal to X; through F draw FG parallel to AB, I. 11 I. 3 I. 31 and through C, D, B draw CH, DK, BG parallel to AF. Proof. Because the rectangle AG is equal to the sum of the rectangles AH, CK, DG; and AG is equal to the rectangle contained by X, AB, for it is contained by AF, AB; and AF is equal to X; Constr. and AH is equal to the rectangle contained by X, AC, for it is contained by AF, AC; and AF is equal to X; Constr. and CK is equal to the rectangle contained by X, CD, for it is contained by CH, CD; and CH is equal to AF, which is equal to X; I. 34 and similarly DG is equal to the rectangle contained by X, DB. Therefore the rectangle contained by X, AB is equal to the sum of the rectangles contained by X, AC, by X, CD, and by X, DB. Q. E. D. Prop. 1 may also be expressed briefly thus: (1) X.AB = X.AC+X.CD + X.DB; or (2) x (a + b + c) = xα + xb + xc, where x is supposed to denote a straight line containing x units of length, and a, b, c lines containing a, b, c units respectively. The sum of the perpendiculars let fall from any point within any equilateral figure upon its sides is the same, wherever the point be taken. PROP. 2.-Theorem.-If a straight line be divided into any two parts, the square on the whole line is equal to the sum of the rectangles contained by the whole line and each of the parts. Let the straight line AB be divided into any two parts AC, CB; then the square on AB shall be equal to the sum of the rectangles contained by AB, AC, and by AB, CB. I. 46 I. 31 Construction. On AB describe the square ADEB, and AE is the square on AB; Constr. and AF is equal to the rectangle contained by AB, AC, for it is contained by AD, AC; and AD is equal to AB; and CE is equal to the rectangle contained by AB, CB, for it is contained by BE, CB; and BE is equal to AB. Therefore the square on AB is equal to the sum of the rectangles contained by AB, AC and by AB, CB. Q. E. D. Prop. 2 may also be expressed thus: (1) or (2) AB2 = AB. AC + AB.CB; (a + b)2 = (a + b)a + (a + b)b, where a and b denote the two parts of a line which contains a+b units of length. PROP. 3.-Theorem.-If a straight line be divided into any two parts, the rectangle contained by the whole line and one of the parts, is equal to the square on that part together with the rectangle contained by the two parts. Let the straight line AB be divided into any two parts AC, CB; then the rectangle contained by AB, AC shall be equal to the square on AC together with the rectangle contained by AC, CB. Construction. -On AC describe the square ADEC, and through B draw BF parallel to AD or CE to meet DE produced in F. I. 46 1. 31 Proof. Because the rectangle AF is equal to the sum of the rectangles AE, CF; and AF is equal to the rectangle contained by AB, AC, and CF is equal to the rectangle contained by AC, CB, for it is contained by CE, CB; and CE is equal to AC, Therefore the rectangle contained by AB, AC is equal to the square on AC together with the rectangle AC, CB. QE. D. Prop. 3 may also be expressed thus: (1) AB.AC AC2 + AC.CB; = or (2) (a + b) a = a2 + ab, where a and b denote the two parts of a line which contains a + b units of length. PROP. 4.-Theorem.-If a straight line be divided into any two parts, the square on the whole line is equal to the sum of the squares on the two parts, together with twice the rectangle contained by the two parts. Let the straight line AB be divided into any two parts AC, CB; then the square on AB shall be equal to the sum of the squares on AC, CB, together with twice the rectangle AC, CB. Construction.-On AB describe the square ADEB; join BD; 1.46 through C draw CFG parallel to AD or BE, meeting BD in F; and through F draw HFK parallel to AB or DE. Proof. I. 31 (1) It is necessary to prove that CK is a square. Because CG and AD are parallel, and BD meets them, Constr. therefore the exterior angle CFB is equal to the interior angle ADB ; opposite I. 29 1. 5 but the angle ADB is equal to the angle ABD ; for AB is equal to AD, both being sides of a square; therefore the angle CFB is equal to the angle ABD, that is to CBF, therefore the side CB is equal to the side CF. I. 6 And the opposite sides of the parallelogram CK are equal; I. 34 therefore CK is equilateral. And the angle CBK is a right angle; Constr. therefore CK is a square described on CB. I. 46, Cor.; Def. 32 Similarly HG is a square described on HF, which is equal to AC. (2) Because the complement AF is equal to the complement FE, I.43 and AF is equal to the rectangle AC, CB; for CF is equal to CB, therefore AF, FE are together equal to twice the rectangle AC, CB. To each of these equals add HG, CK which are equal to the squares on AC, CB; therefore HG, CK, AF, FE are together equal to the squares on AC, CB, together with twice the rectangle AC, CB. But HG, CK, AF, FE make up AE, which is the square on AB; therefore the square on AB is equal to the sum of the squares on AC, CB together with twice the rectangle AC, CB. Q. E. D. Corollary. Parallelograms about a diagonal of a square are also squares. ALTERNATIVE PROOF. PROP. 4.-Theorem.-If a straight line be divided into any two parts, the square on the whole line is equal to the sum of the squares on the two parts, together with twice the rectangle contained by the two parts. Let the straight line AB be divided into any two parts AC, CB; then the square on AB shall be equal to the sum of the squares on AC, CB, together with twice the rectangle AC, CB. Proof. -Because AB is divided into two parts at C, therefore by Prop. 2 the square on AB is equal to the sum of the rectangles AB, AC and AB, CB. But by Prop. 3 the rectangle AB, AC is equal to the square on AC, together with the rectangle AC, CB; and also by Prop. 3 the rectangle AB, CB is equal to the square on CB, together with the rectangle AC, CB. Therefore the square on AB is equal to the sum of the squares on AC, CB, together with twice the rectangle AC, CB. Prop. 4 may be expressed thus: (1) AB2 = AC2 + CB2 + 2 AC.CB; Q. E. D. or (2) (a + b)2 = a2 + b2 + 2 ab, where a and b denote the two parts of a line which contains a + b units of length. |