234 MATHEMATICS. SECOND STAGE. SECTION I. GEOMETRY. EUCLID'S ELEMENTS, BOOK II. Definitions. 1. A rectangle, or right-angled parallelogram, is said to be contained by any two of the straight lines which contain one of the right angles. 2. In any parallelogram, the figure which is composed of either of the parallelograms about a diameter, together with the two complements, is called a gnomon. Thus the parallelogram HG, together with the complements AF, FC, is a gnomon, which is briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon. Α E F H B G D The rectangle under, or contained by two lines, as AB and BC, is concisely expressed thus:-AB, BC. Proposition 1.-Theorem. If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E; ABD + The rectangle contained by the straight lines A and BC A·BC= shall be equal to the rectangle contained by A and BD, together with that contained by A and DE, and that contained by A and EC. CONSTRUCTION. From the point B draw BF at right angles to BC (I. 11), And make BG equal to A (I. 3). Through G draw GH parallel to BC (I. 31), And through the points D,E,C,draw DK, EL, CH parallel to BG (I. 31). PROOF. Then the rectangle BH is equal to the rectangles BK, DL, EH. B D F с K A H A EC. Since BH = BK + DL + But BH is contained by A and BC, for it is contained EH. by GB and BC, and GB is equal to A (Const.); And BK is contained by A and BD, for it is contained by GB and BD, and GB is equal to A ; And DL is contained by A and DE, because DK is equal to BG, which is equal to A (I. 34) ; And in like manner EH is contained by A and EC; Therefore the rectangle contained by A and BC is equal to the several rectangles contained by A and BD, by A and DE, and by A and EC. Therefore, if there be two straight lines, &c. Q. E. D. Proposition 2.-Theorem. If a straight line be divided into any two parts, the rectangles contained by the whole line and each of its parts are together equal to the square on the whole line. Let the straight line AB be divided into any two parts in the point C; AB⚫BC For AB2 is the sum of its The rectangle contained by AB and BC, together with the rectangle contained by AB and AC, shall be equal to the square on AB. CONSTRUCTION.-Upon AB describe the Through C draw CF parallel to AD or PROOF. Then AE is equal to the rect- But AE is the square on AB; A D B Therefore the square on AB is equal to the rectangles AF parts AF+ and CE. CE. And..= And AF is the rectangle contained by BA and AC, for it is contained by DA and AC, of which DA is equal to BA; And CE is contained by AB and BC, for BE is equal to AB. Therefore the rectangle AB, AC, together with the rect AB AC+ angle AB, BC, is equal to the square on AB. Therefore, if a straight line, &c. Q. E. D. AB BC. AB BC= AC BC. For AE AD +CE. Proposition 3.-Theorem. If a straight line be divided into any two parts, the rect angle contained by the whole 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 in the point C; The rectangle AB BC shall be equal to the square on BC, together with the rectangle AC CB. CONSTRUCTION.-Upon BC describe the (I. 46). Produce ED to F; and through A draw AF parallel to CD or BE (I. 31). PROOF. Then the rectangle AE is equal to the rectangles AD and CE. But AE is the rectangle contained by AB and BC, for it is contained by AB and BE, of which BE is equal to BC; square CDEB D And AD is contained by AC and CB, for CD is equal to CB; And CE is the square on BC. Therefore the rectangle AB, BC is equal to the square on BC, together with the rectangle AC, CB. Therefore, if a straight line, &c. Q. E. D. Proposition 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 parts. Let the straight line AB be divided into any two parts in C ; The square on AB shall be equal to the squares on AC and AB2 = CB, together with twice the rectangle contained by AC AC2+CB2 and CB. CONSTRUCTION.-Upon AB describe the square ADEB (I. 46), and join BD. Through C draw CGF parallel to AD or BE (I. 31). Through G draw HGK parallel to AB H or DE (I. 31). PROOF.-Because CF is parallel to AD, and BD falls upon them, Therefore the exterior angle BGC is D equal to the interior and opposite angle ADB (I. 29). B K Because AB is equal to AD, being sides of a square, the angle ADB is equal to the angle ABD (I. 5); +2AC CB. Therefore the angle CGB is equal to the angle CBG Show first (Ax. 1); Therefore the side BC is equal to the side CG (I. 6). It is likewise rectangular. For since CG is parallel to BK, and CB meets them, the angles KBC and GCB are together equal to two right angles (I. 29). But KBC is a right angle (Const.), therefore GCB is a right angle (Ax. 3). Therefore also the angles CGK, GKB, opposite to these, are right angles (I. 34). that CK is So also Therefore CGKB is rectangular; and it has been proved equilateral; therefore it is a square; and it is upon the side CB. For the same reason HF is also a square, and it is on HF AC2 the side HG, which is equal to AC (I. 34). And Therefore HF and CK are the squares on AC and CB. And because the complement AG is equal to the complement GE (I. 43), And that AG is the rectangle contained by AC and CG, =2AC CB. that is, by AC and CB, ... whole figure or AD DB +CD2 =CB2. Therefore GE is also equal to the rectangle AC, CB; Therefore AG, GE are together equal to twice the rectangle AC, CB; And HF, CK are the squares on AC and CB. Therefore the four figures HF, CK, AG, GE are equal to the squares on AC and CB, together with twice the rectangle AC, CB. But HF, CK, AG, GE, make up the whole figure ADEB, which is the square on AB; Therefore the square on AB is equal to the squares on AC and CB and twice the rectangle AC CB. Therefore, if a straight line, &c. Q. E. D. COROLLARY.-From this demonstration it follows that the parallelograms about the diameter of a square are likewise squares. Proposition 5.-Theorem. If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line. Let the straight line AB be bisected in C, and divided unequally in D; The rectangle AD, BD, together with the square on CD, shall be equal to the square on CB. CONSTRUCTION. Upon CB describe the square CEFB (I. 46), and join BE. |