Similarly if AB contains a units, and AD, b units, then the area of AC will be ab square units. From this we see that if lines can be represented by numbers, the area of the rectangles formed by them will be represented by the products of those numbers. It is this connection between lines and numbers, areas and products, which leads to propositions in Arithmetic and Algebra corresponding to the first ten propositions of this book. But algebraical proofs, as they are called, may not be substituted for the geometrical ones for these reasons: 1. Because these methods of proof can only apply when the lines are commensurable, whereas the propositions are true for all divisions of the lines, whether commensurable or not. 2. Because in Arithmetic and Algebra we proceed by general rules and processes without stopping to interpret each step of the operation. But one of the principal advantages of the study of geometry is that each truth is shown in its connection with what has gone before, and we are made to understand every step taken. If, therefore, the methods of the former were substituted for the latter, one of the principal advantages to be derived from a study of geometry would be lost. The methods of Algebra apply to Algebraical problems, and the methods of Geometry should be retained in dealing with Geometry. For these reasons also the symbols of Algebra should not be used in Geometry. BOOK II. DEFINITIONS. 1. Every right-angled parallelogram or rectangle is said to be contained by any two of the straight lines which contain one of the right angles. 2. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a gnomon. Thus the parallelogram A E HG, together with the complements AF, FC, is the gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite H angles of the parallelograms which make the B gnomon. PROPOSITION 1. THEOREM. K 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. 1. E: Let A and BC be two straight lines; and let BC be divided into any parts at the points D, the rectangle contained by the straight lines A, BC, shall be equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC. D E 2. 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; and through D, E, C draw DK, EL, CH, parallel to BG. [I. 31.] 3. Then the rectan gle BH is equal to A the rectangles BK, DL, EH. But BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A. [Const.] And BK is contained by A, BD, for it is contained by GB, BD, and GB is equal to A; and DL is contained by A, DE, because DK is equal to BG, which is equal to A; [I. 34.] and in like manner EH is contained by A, EC. Therefore the rectangle contained by A, BC is equal to the several rectangles contained by A, BD, and by A, DE, and by A, EC. Therefore, if there be two straight lines, &c. Q.E.D. The corresponding proposition in Arithmetic is the foundation of the rule for simple multiplication, and may be thus expressed. If there be two numbers, one of which is divided into any number of parts, the product of these numbers is equal to the sum of the partial products obtained by multiplying by the second number each of the parts into which the first number is divided. Thus 385 x 7=300 x 7 +80 × 7+5x7 or more generally, PROPOSITION 2. THEOREM. If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square on the whole line. 1. Let the straight line AB be divided into any two parts at the point C: the rectangle contained by AB, BC, together with the rectangle AB, AC, shall be equal to the square on AB. 2. On AB describe the square A ADEB; and through C draw CF parallel to AD or BE. [I. 31.] 3. Then AE is equal to the rectangles AF, CE. But AE is the square on AB. And AF is the rectangle contained by BA, AC, for it is contained by D DA, AC, of which DA is equal to BA; and CE is contained by AB, BC, for BE is equal to AB. Therefore the rectangle AB, AC, together with the rectangle AB, BC, is equal to the square on AB. Therefore, if a straight line, &c. Q.E.D. Ex. Put into words the arithmetical proposition corresponding to this, and if m=a+b, shew that its symbolical expression is m2=am+bm. PROPOSITION 3. THEOREM. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square on the aforesaid part. 1. Let the straight line AB be divided into any two parts at the point C: the rectangle AB, BC shall be equal to the rectangle AC, CB, together with the square on BC. 2. On BC describe the A square CDEB; and produce ED to F; and through A draw AF parallel to CD or BE. [1.31.] Then the rectangle 3. AE is equal to the rectangles AD, CE. But AE is the rectangle con E tained by AB, BC, for it is contained by AB, BE, of which BE is equal to BC; and AD is contained by AC, CB, for CD is equal to CB; and CE is the square on BC. Therefore the rectangle AB, BC is equal to the rectangle AC, CB, together with the square on BC. Therefore, if a straight line, &c. Q.E.D. If AC-m and CB= Ex. Construct the figure and prove the proposition so as to show that the rectangle AB, AC is equal to the rectangle AC, CB together with the square on AC. PROPOSITION 4. THEOREM. If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the two parts. 1. Let the straight line AB be divided into any two parts at the point C: the square on AB shall be equal to the squares on AC, CB, together with twice the rectangle contained by AC, CB. On AB describe the square 2. ADEB; and join BD; and through C draw CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. [I. 31.] Then, because CF is par 3. allel to AD, and BD falls on them, the exterior angle BGC H is equal to the interior and oppo- D F B E site angle ADB; [I. 29.] but the angle ADB is equal to the angle ABD, [I. 5.] because BA is equal to AD, being sides of a square; therefore the angle CGB is equal to the angle CBG; [Ax. 1.] and therefore the side BC is equal to the side CG. [I. 6.] But CB is also equal to GK, and CG to BK; [I. 34.] therefore the figure CGKB is equilateral. It is likewise rectangular. For since CG is parallel to BK, and CB meets them, the angles KBC, GCB are together equal to two right angles. [I. 29.] But |