COROLLARY 3; PROBLEM. To construct a square equal in area to the sum of two or more given squares. SOLUTION. Let the straight lines A, B, and C be sides of the given squares. Draw ED and DF at right angles, and make GD equal to A and DH equal to B; join GH, and draw GI perpendicular to it; make GK equal to C, and join KĤ, then a square constructed on the line KH shall be the square required. DEMONSTRATION. For the square on KH is equal in area to the sum of the squares on GK or C, and on GH (a), but the square on GH is equal in area to the squares on GD and DH, or on A and B (a); therefore the square on KH is equal in area to the sum of the squares on A, B, and C. (a) I. 47. COROLLARY 4; PROBLEM. To construct a square equal in area to the difference of two given squares. SOLUTION. Let A and B be sides of the given squares. Draw CD equal to the side of the greater square B, and produce it until the produced part DE is equal to the side of the other square A. From the center D, at the distance CD, describe the circle CFG, and through E draw EF perpendicular to CE, to meet the circle in F; then FE is a side of the required square. DEMONSTRATION. Join DF. The square on DF, or its equal B, is equal in area to the sum of the squares on DE or A, and on EF (a); therefore if the square on DE (a) I. 47. be taken from the square on DF, the difference is equal in area to the square on EF. &c. COROLLARY 5; PROBLEM. To find geometrical values of 1, 2, √3, SOLUTION. Take AB equal to 1; draw CB perpendicular to AB, and also equal to 1; join AC, and draw DC perpendicular to it and equal to 1; join DA, and draw DE perpendicular to it and equal to 1, fc.; then the lines AC, AD, AE, &c., are respectively equal to 1, 2, 3, &c. DEMONSTRATION. The square on AC equals the sum of the squares on AB and BC (a); the squares on AB and BC each equal 1 (b); therefore the square on AC equals 2, and AC equals 2. Again, the square on AD equals the sum of the squares on AC and DC (a); the square on AC equals 2, and the square on A (a) I. 47. DC equals 1 (b); therefore the square on AD equals 3, and AD equals 3. Therefore the square roots of the natural numbers 1, 2, 3, 4, &c., are represented geometrically by the lines AB, AC, AD, AE &c. D COROLLARY 6. By means of the 47th proposition, any two sides of a rightangled triangle being given, the third side may be readily found; for if the two sides given are those which form the right angle, the sum of the squares on those two sides will be equal to the square on the third side; and if one of the given sides subtends the right angle, then the square on the third side is equal to the difference of the squares on the two given sides. This application of the 47th proposition to the purposes of finding the parts of right-angled triangles is of immense use, and is really the foundation of Trigonometry. COROLLARY 7. The 47th proposition holds true, if, instead of squares, we construct any similar figures on the sides of the triangle, such as circles, equilateral triangles, &c., and may therefore be generalized as follows: THEOREM. If a triangle be right-angled, any figure. which is constructed upon the side subtending the right angle is equal in area to the sum of the similar figures constructed upon the sides which form the right angle. SCHOLIUM. If three semicircles be described on the three sides of a right-angled triangle, the area of ABC will therefore be equal to the sum of the areas of ADB and BFC; if, now, from each we deduct the common segments AEB and BGC, we have the triangle ABC equal in area to the sum of two lunes ADBE and BFCG. This proposition D was discovered by Hippocrates, and was the first instance of the determination geometrically of the area of a space entirely bounded by curved lines; it led him to believe that, by means of this proposition, he would be able to solve the problem, to determine geometrically the area of the circle. PROPOSITION XLVIII. THEOREM.-If the square constructed upon one side (BC) of a triangle (ABC) be equal in area to the sum of the squares constructed upon the other two sides (AC and AB), the angle BAC opposite to that side is a right angle. CONSTRUCTION. From the point A draw AD at right angles to one of the sides AC (a), and equal to the other AB (b); and join DC. DEMONSTRATION. Because DA equals AB (c), therefore the square on DA equals the square on AB (d); and adding to each the square on AC, therefore the sum of the squares on DA and AC is equal to the sum of the squares on AB and AC (e). But the square on DC is equal in area to the sum of the squares on DA and AC (f), because DAC is a right angle (c); and the square on BC is equal in (a) I. 11. (b) I. 2. (c) Const. (d) I. 46, cor. 1. (e) Ax. 2. (f) I. 47. area to the sum of the squares on AB and AC (g); therefore the square on DC is equal to the square on BC, and therefore DC is equal to BC (h). Then, because in the triangles DAC and BAC the sides AB and AD are equal, AC common to both, and the base DC equal to BC, therefore the angle DAC is equal to BAC (); but DAC is a right angle (c), therefore BAC is also a right angle. SCHOLIUM. This proposition is the converse of the preceding one. (c) Const. THE ELEMENTS OF EUCLID. BOOK II. DEFINITIONS. 1. EVERY rectangle is said to be contained by any two of the straight lines which contain one of the right angles. SCHOLIUM. As already explained in the scholium to the twenty-seventh definition in the former Book, a rectangle is designated as the rectangle under the two lines by which it is contained. 2. In any parallelogram, either of the parallelograms about the diagonal, together with the two complements, is called a gnomon. THEOREM.-If there be two straight lines (A and BC), one of which is divided into any number of parts (BD, DE, EC), the rectangle under the two lines is equal in area to the sum of the rectangles under the undivided line (A) and the several parts of the divided line (BD, DE, EC). CONSTRUCTION. From the point B draw BF perpendicular to BC (a), and make BG equal to A (b); through G draw GH parallel to BC (c), and through D, E, C, draw DK, EL, CH, parallel to BG (c). F (a) I. 11. I. 3. DEMONSTRATION. It is evident that the rectangle BH is equal to the sum of the rectangles BK, DL, EH; but the rectangle BH is the rectangle under A and BC, for BG is equal to A (d); and the rectangles BK, DL, ÉH, are respectively the rectangles under A and BD, A and DE, A and EC, for each of the lines BG, DK, EL, is equal to A (e). Therefore the rectangle under (c) I. 31. A and BC is equal in area to the sum of the rectangles under ▲ and BD, A and DE, and A and EC. A D C E B + SCHOLIUM. It is of considerable importance that the true relationship or connection between Geometry and Analysis be clearly understood before entering upon the second book of the Elements. The subject of Geometry is magnitude; that of Analysis, i. e., Arithmetic and Algebra, is number and quantity. In order, therefore, to a just understanding of the connection between Geometry and Algebra, we must first obtain a clear notion of the connection between the subject of each, that is, between magnitude and number. Let us take a line AB of any given magnitude, and bisect it in C; and let us again bisect AC in D, and CB in E. We thus obtain four equal lines, which are together equal to the given line AB; and it is evident that, being all equal, any one of those lines taken four times will be equal to the four lines, or to the original line AB; and we have thus two modes of expressing the magnitude of the line AB, namely, either by saying that it is equal to four lines each equal to AD, or by saying that it equals AB. That is, we may either view the line in its entirety, or we may-having first conceived it as being divided into any number of equal parts-view it as the magnitude of one of those parts repeated that number of times. Each of these parts is termed a unit; the process which we follow in order to determine the number of units in a given line is termed measuring that line; and if the unit is found to be contained any exact number of times in the given line, that is to say, if the unit added on to itself any certain definite number of times forms a line neither longer nor shorter than the given line, but exactly coinciding therewith, the unit is termed the measure of the line; and if the same unit is found to measure any other given line, so that being taken any other certain definite number of times, it forms another line exactly equal to the second given line, that unit is said to be the common measure of both the given lines, and those two lines are said to be commensurable. A unit may be arbitrarily determined, that is to say, we may assume a line of any length that we please for the purpose; but having once determined the magnitude that shall constitute the unit, that magnitude must be considered as fixed and unalterable. Thus, in any particular course of investigation, we may assume for the unit a line a yard in length, or a foot in length, or an inch in length; and having done so, we should express the magnitude of any line by the number of lines (each one yard, foot, or inch, as the case might be) which would form a line equal in magnitude to the given line. Here, then, we see the connection between number and magnitude; number may be regarded as the instrument through the medium of which we estimate and express magnitude. Were we not in possession of the common notion or idea of number, we could only express the magnitude of any given line by the actual exhibition of the line, or of another equal to it; but, having that idea, we are enabled to declare its magnitude by comparing it with another standard line (termed a unit), the magnitude of which is already familiar to us; and we thus make known the magnitude of the given line, by stating the result of that comparison, or the number of those units which the line is equal to. We have hitherto assumed that the given line has, in every case, been equal to a certain number of units; but let us now suppose that, having arbitrarily fixed upon a unit, when we apply it to the given line, as AB, by cutting off successive portions AC, CD, DE, &c., each equal to the unit, we at length arrive at a remaining portion, as FB, which is less than the unit; how are we, in such a case, to determine the magnitude of the given line? The most obvious mode 早早 |