F E K A D G M м B H A L be in the same straight line with AB, * and produce FG to H. Through A draw (1. 31.) AH parallel to BG or EF, and join HB. Then, because HF falls upon the parallels AH, EF, the angles AHF, HFE, are together equal (I. 29.) to two right angles ; wherefore the angles BHF, HFE are less than two right angles; and therefore (I. ax. 12.) HB, FE will meet, if produced : let them meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB to the points L, M. Then FL is a parallelogram, of which the diagonal is HK, and AG, ME are the parallelograms about HK; and LB, BF are the complements; therefore LB is equal (I. 43.) to BF: but BF is equal to the triangle C; wherefore LB is equal (I. ax. 1.) to C; and because the angle GBE is equal (1. 15.) to À BM, and likewise to D, the angle ABM is equal (I. ax. 1.) to D. Therefore the parallelogram LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to D: which was to be done. PROP. XLV. PROB. F GL To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given angle. Let ABCD be a given rectilineal figure, and E a given angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Join DB, and describe (I. 42.) the parallelogram FH equal to the triangle ADB, and having the angle K equal to E. Produce KH to M, and to the straight line GH apply (I. 44.) the parallelogram GM equal to the triangle DBC, having GHM for one of its angles : and because the straight line HG meets the parallels KM, FL, the alternate angles MHG, HGF are equal : add to each of these the angle HGL: therefore the angles MHG, HGL are equal to HGF, HGL: but MHG, HGL are equal (I. 29.) to two right angles ; wherefore also HGF, HGL are equal to two right angles, and FG is therefore (I. 14.) in the same straight line with GL. Because also KF is parallel to HG, and HG to ML; KF is parallel (I. 30.) to ML: and KM, FL are parallels; wherefore (I. def. 24.) KL * In strictness, a parallelogram should be constructed (1. 42.) equal to the triangle C, and having an angle equal to D. Then, AB being produced, the parallelogram BF should be constructed, by propositions 23, 3, and 31, having the angle E BG equal to D, and the sides EB, BG respectively equal to two adjacent sides of the parallelogram previously constructed : and this parallelogram being equal (I. ax. 1. and I. 34. cor. 3.) to the triangle C, the rest of the construction will proceed as in the text. is a parallelogram. And because the triangle ABD is equal to the parallelogram RF, and the triangle DBC to the parallelogram GM; the whole figure ABCD is equal to the whole parallelogram KL. Therefore the parallelogram KL has been described equal to the given rectilineal figure ABCD, and having the angle K equal to the given angle E: which was to be done. Cor. From this it is manifest how to a given straight line a parallelogram may be applied, which shall have an angle equal to a given angle, and shall be equal to a given rectilineal figure; viz., by applying (I. 44.) to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle. The rest of the construction will be the same as in the proposition. PROP. XLVI. PROB. D B E A B В To describe a square upon a given straight line. Let AB be a given straight line; it is required to describe a square upon it. From the point A draw (1. 11.) AC at right angles to AB, and make AD equal to AB: through * the point D draw (1.31.) DE parallel to AB, and through B draw BE parallel to AD. Then (I. def. 26. and const.) the figure AE is a rectangle; and since AD, AB are equal, AE is (I. def. 27.) a square; and being described on the given line AB, it is the square required. Cor. 1. Hence (const. and I. 34.) a square has all its sides equal: and (I. 34. cor. 1.) all its angles are right angles. Cor. 2. Hence the squares described on equal straight lines (1. 34. cor. 3.) are equal. Cor. 3. If two squares be equal, their sides are equal. For, if the sides be not equal, from one of them cut off a part equal to the other, and on that part describe a square. This square (by the preceding corollary, and I. ax. l.) will be equal to the square of which it is a part, which is absurd. Cor. 4. If AB and AD, two adjacent sides of a rectangle BD, be divided into parts AE, EF, AH, &c., which are all equal; straight lines drawn through E, F, H, &c., the points of section, parallel to the sides, divide the rectangle into squares, HE, FL, KL, &c., which are all equal, and the number of which is equal to the product of the number of parts in AB, one of the sides, multiplied by the number of parts in E F G B H K D с The part of the construction that follows is effected more simply in practice, by describing arcs from B and D as centres, and with radii equal to AB, as these will intersect each other in E. In the construction given in the text, it should be proved that DE and BE will meet. This is easily done by ineans of the 12th axiom, is B D be joined. The student will perceive that there inight be another square described on AB, on the other side. AD, the other. For (by this proposition and its 2d corollary, and by the 34th and 29th of this book,) HE, FL, &c., are squares, and are all equal; and of these squares there are evidently as many columns as there are parts in AB, while in each column there are as many squares as there are parts in AD. The number of such squares contained in a figure is called, in the language of mensuration, the area of that figure. Cor. 5. Hence, since any parallelogram is equal (I. 35.) to a rectangle on the same base and between the same parallels, it follows, that the area of any parallelogram is equal to the product of its base and its perpendicular height: and the area of a square is computed by multiplying a side by itself. Cor. 6. Hence, also (I. 41.) the area of a triangle is computed by multiplying any of its sides by the perpendicular drawn to that side from the opposite angle, and taking half the product : and the area of a trapezium is found by multiplying either diagonal by the sum of the perpendiculars drawn to it from the angles which it subtends, and taking half the product. Cor. 7. Every polygon may be divided into triangles or trapeziums by drawing diagonals ; and, therefore, the area of any polygon whatever can be computed by finding the areas of those component figures, by the last corollary, and adding them together.t PROP. XLVII. THEOR. In any right-angled triangle, the square which is described upon the hypotenuse, that is, the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle. Let ABC be a right-angled triangle, having the right angle BAC; the square described upon the hypotenuse BC is equal to the squares described upon BA, AC. On BC describe (I. 46.) the square BE, and on BA, AC, the Thus, if the length of a rectangle be 4 feet, and its breadth 3 feet, it will contain 12 squares, the side of each of which is one foot; or, as it is briefly expressed, its area is 12 square feet: and a rectangle 15 inches long, and 10 inches broad, contains 150 square inches ; that is, 150 squares, each having its side an incb. † This corollary and the two foregoing contain the elementary principles of the mensuration of rectilineal figures, and they form a connexion between arithmetic or algebra, and geometry. They also explain the origin of the expressions, “ the square of a number," " the rectangle of two numbers," and “the product of two lines.” The last of these is a convenient abbreviation for the product of the number of equal parts in one line by those in another. The second is not only an improper, but an unnecessary substitute for “the product of two numbers.” Instead of the first, it would be better to employ “the second power of a number;" but the expression is in so general use that it is not likely to be altogether laid aside. G H K B с D LE squares GB, HC; through A draw (I. 31.) AL parallel to BD, or CE, and join AD, FC. Then, because each of the angles BAC, BAG is a right angle, the two straight lines AC, AG upon the opposite sides of AB, make with it, at the point Á, the adjacent angles equal to two right angles ; therefore CA is in the same straight line (I. 14.) with AG. For the same reason, AB and AH are in the same straight line. Again, because the angle DBC is equal to FBA, each of them being a right angle, add to each the angle ABC, and the whole angle DBA is equal (1. ax. 2.) to the whole FBC: and because the two sides AB, BD are equal (const.) to the two FB, BC, each to each, and the angle DBA equal to FBC; therefore the triangle ABD is equal (I. 4. part 2.) to the triangle FBC. Now the parallelogram BL is double (I. 41.) of the triangle ABD, because they are upon the same base BD, and between the same parallels BD, AL; and the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal (I.ax. 6.) to one another: therefore the parallelogram BL is equal to the square GB. In the same manner, by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the square HC. Therefore the whole square BE is equal (I. ax. 2.) to the two squares GB, HC; and the square BE is described upon the hypotenuse BC, and the squares GB, HC upon BA, AC: wherefore the square upon BC is equal to the squares upon BA, AC.* Therefore, in any right-angled triangle, &c. Cor. 1. Hence, if the lengths of any two sides of a right-angled triangle be given in numbers, the remaining side can be found. + Cor. 2. Hence, also, we may find a square equal to the sum of two or more given squares. Thus, let AB be the side of one square, and AC perpendicular to it, the side of another, and join BC: the square of BC is equal to the sum of the squares of AB and AC. In like manner, if CD be drawn perpendicular to BC, and equal to the * In the proof here given, which is as simple and easy as any other that has been proposed instead of it, the squares are all described externally in respect to the triangle. This, however, is not essential, as the squares may be described either all on the other sides of the lines AB, AC, BC, or some on one side and some on the other. The proof is nearly alike in all these modes, the chief difference consisting in using, in some instances, instead of the 4th proposition, the 2d corollary to the 38th. + Thus, if AB be 15 inches, and AC 8 inches, the areas of the squares BG and CH are (I. 46. cor. 5.) 225 and 64 square inches. The sum of these, 289, is, by this proposition, the area of BE; the square root of which is 17 inches, the length of BC. In like manner, if BC be 13 percbes, and AC 5 perches, the areas of BE and CH are 169 and 25 square perches ; the difference of which, 144, is, by this proposition, the area of BG; whence, by extracting the square root, AB is found to be 12 perches. D с B A B А B D C B side of a third square, and if BD be joined, the square of BD is equal to the squares of BC, CD, or to those of BA, AC, CD: and thus a square may be found equal to the sum of any number of given squares. Cor. 3. A square may also be found equal to the difference of two given squares. Thus, to AB the side of the less square draw the perpendicular BC; and from A as centre, with the side of the greater as radius, describe an arc cutting the perpendicular in C: the square of BC will be equal to the difference of the given squares. Cor. 4. If a perpendicular AD be drawn from one of the angles of the triangle A BC, cutting the opposite side in D, the difference of the squares of the sides AB, AC, is equal to the difference of the squares of the seg ments BD, CD, intercepted between the extremities of the base and the point in which the perpendicular cuts it. For the square of AB is equal to the squares of BD and DA, and the square of AC to the squares of DC and DA; and by taking the latter equals from the former, there remains the difference of the squares of AB and AC equal to the difference of the squares of the segments BD and CD.* Cor. 5. If two right-angled triangles ABC, DEF have their hypotenuses AB, DE equal, and likewise the sides BC, EF equal ; the remaining sides AC, DF are also equal; and the angle A is equal to D, and B to E. For, since the hypotenuses are equal, the squares of BC and CA are equal to the squares of EF and FD; of which also the squares of BC and EF are equal; and therefore (I. ax. 3.) the squares of AC and DF, and consequently (I. 46. cor. 3.) AC and DF themselves are equal. Also, since the sides of the triangles are respectively equal, their angles (I. 8.) are likewise equal, each to each. The converse of this, that if the legs t of two right-angled triangles be equal, each to each, their hypotenuses are also equal, may be proved in a similar manner, or by means of the 4th proposition of this book. Cor. 6. If the hypotenuses AB, DE be equal, but BC, a leg of one of them, be greater than EF, a leg of the other, the remaining leg AC of the first triangle is less than DF, the 44 In this proof the obvious principle is employed, that the difference of two magnitudes is the same as the difference obtained after adding to each the same third magnitude. Thus, the difference of the squares of BD and DC is the same as the difference between the sum of the squares of BD and DA, and of DC and DA. + The sides containing the right angle are often called the legs of the triangle. |