А. В CDA B D C DB B will be equal to the rectangle under the Let A B C be any segments of the hypotenuse; and that, triangle, and C one conversely, if this be the case, the angle of its angles ; and BAC will be a right angle. (See the from the angle A to proof of Cor. 2.) the opposite side BC, or B C pro- duced, let there be Among the different proofs which drawn the perpenhave been invented of this celebrated dicular A D: the theorem, there is one of no little ele- square of A B shall be greater or less gance, which has the advantage of than the squares of AC, C B, by twice pointing out in what manner the squares the rectangle BC, CD; greater, if C of the two sides may be dissected, so as be greater than a right angle, and less to form, by juxta-position of their parts, if it be less. the square of the hypotenuse. It will When C is greater than a right angle, be readily apprehended from the follow- the opposite side, A B, and the perpendiing outline. BAC is cular, AD), must lie upon different sides a triangle, having the of A C; since, otherwise, in the triangle angle at A a right ACD, one of the angles would be a angle: upon the hypo right angle, and another greater than a tenuse B C is described right angle, which is impossible (8.). For the square BCDE, the like reason, when C is less than a and through the points i right angle, the opposite side, A B, and D, E, straight lines are the perpendicular, AD, must lie upon drawn parallel to B A, C A, to meet one the same side of AC. another in K, and AC, A B produced, Therefore, according as C is greater in the points F, G. Then it may be or less than a right angle, the line BD shewn that the four right-angled trian- will be the sum, or the difference of gles, upon the sides of the square BD, BC, CD and (32. and 33.) the square are equal to one another; and, there- of B D will be greater or less than the fore together equal to twice the rectan- squares of BC, C D by twice the rectgle BA, AC: also, that the figure angle BC, C D. Add to each the AFKG is a square, and equal to the square of AD: therefore, the squares square of B A+AC: and hence it of B D, A D will be greater or less than easily follows that the square of B C is the squares of B C, Č D, A D, by twice equal to the squares of B A, AC(32. and the rectangle B C, C D. But the square ax. 3.). To shew the dissection of the of A B is equal (36.) to the squares of squares, Ee and Cc are drawn paral- BD, AD, and the square of A C is lel to A B, to meet B b, which is drawn equal to the squares of CD, A D. parallel to A C, in the points c, e: then Therefore, the square of A B will be Cb is equal to the square of A B, and greater or less than the squares of BC, Eb to the square of A C; and the former AC by twice the rectangle BC, CD. is divided into two, and the latter into Therefore, &c. three parts, which may be placed (as Cor. Any angle of a triangle is greater indicated by the divisions of B D,) so as or less than a right angle, according as to fill up the space BCDE, which is the square of the side opposite to it is the square of BC. greater or less than the squares of the sides by which it is contained. PROP. 37. (Euc. ii. 12. and 13.) Scholium. In every triangle, the square of the When the point D coincides with side which is opposite to any given angle C is a right angle , and we come C, there is no rectangle BC, CD, the angle, is greater or less than the squares to the conclusion of Prop. 36.: when D of the sides containing that angle, by twice the rectangle, contained by either coincides with B, the rectangle BC, CD of these sides, and that part of it, which becomes the square of B C, the angle is intercepted between the perpendicular B is a right angle, and we come to the let fall upon it from the opposite angle, conclusion of Prop. 36. Cor. 1. and the given angle : greater, when the PROP. 38. given angle is greater than a right In every triangle, if a perpendicular angle, and less, when it is less. be drawn from the vertex to the base, B DC B ср А В DE CBDC D or to the base produced ; the difference of the squares of AD, AC; that is, the of the squares of the sides shall be equal square of AD is less or greater than the to the difference of the squares of the square of A C, by the rectangle BD, segments of the base, or of the base pro- D°C. duced. Therefore, &c. Let ABC be a PROP. 40. triangle, and from the vertex A to the In every triangle the squares of the base BC, or BC two sides are together double of the produced, let there squares of half the base, and of the be drawn the per straight line, which is drawn from the pendicular AD: the difference of the vertex to the bisection of the base. squares of AB, A C, shall be equal to the Let ABC be difference of the squares of B D, CD. any triangle, and For the square of B D is as much from the vertex A greater than the square CD, as the to D, the middle squares of B D, A D together are greater point of the base, than the squares of C D, A D together, let there be drawn that is (36.), as the square of A B is the line A D. The greater than the square of A C. squares of A B, A C, shall be together Therefore, &c. double of the squares of B D, DA. We may remark that, if the triangle From the point A to B C, or B C probe isosceles, the segments of the base duced, draw the perpendicular A E (12.). will be equal; and, that, in other cases, Then BE is equal to the sum of BD, DE; the greater segment of the base is always and, because B D is equal to DC, EC adjoining to the greater side. (12. Cor. is equal to the difference of B D, DE. l, and 2.) Therefore, (35.) the squares of B E, EC are together double of the squares of PROP. 39. BD, DE. And the square of EA, taken In an isosceles triangle, if a straight twice, is double of the square of E A. line be drawn from the vertex to any. Therefore the squares of B E, E A, C E, point in the base, or in the base pro- E A, are together double of the squares duced, the square of this straight line of B D, DE, EA; that is (36.), the shall be less or greater than the square squares of BA, AC are together of either of the two sides, by the rect- double of the squares of B D, DA. angle under the segments of the base, or When ACB is a right angle, the of the base produced. perpendicular AE coincides with the Let ABC be side A C. Therefore B D is equal to an isosceles tri D E, and the square of B E is double of angle, having the the squares of BD, DE (29.Cor. 2.); also side AB equal the square of BE (36. Cor. 1.) is equal to the side AC, to the difference of the squares of B A, and from the A E, that is, of B A, AC; and hence vertex A to any point D in the base, or the squares of B A, A E, that is, of B A, in the base produced, let there be drawn AC, are together double of the squares the line AD: the square of AD shall of B D, D A, as before. be less or greater than the square of Therefore, &c. A C, by the rectangle B D, DC. Bisect BC in E, (post. 3.) and join AE. PROP. 41. Then, DC is equal to the sum of DE and The squares of the four sides of a EC; and because B E is equal to EC, quadrilateral are together greater than BD is equal to the difference of D E the squares of its diagonals, by four and E C. Therefore the rectangle BD, times the square of the straight line D C is (34.) equal to the difference of which joins the middle points of the the squares of DE, E C. But the line diagonals. AE is perpendicular to BC, because the Let ABCD be base BC of the isosceles triangle is bi- a quadrilateral, and sected in E (6. Cor. 3.). Therefore, (38.) let its diagonals A C, the difference of the squares of DE,EC, BD be bisected in is equal to the difference of the squares the points F, E: the of AD, AC. Therefore (ax. 1.) the rectsquares of A B, angle BD, DC, is equal to the difference, BC, CD, D A, are BD E DB А B 9 of EF; together greater than the squares of Operations so simple require no further AC, BD, by four times the square of notice. EF. PROP. 42. Prob. 1. (Euc. i. 1.) Join B F, FD. Then, because the base A C of the triangle BAC is bi To describe an equilateral triangle sected in F, the squares of A B, B C are upon a given finite straight line A B. (40.) equal to twice the squares of B F, From the points A, B, AF: and for the like reason the squares as centres, with the comof CD, DA are equal to twice the mon radius A B, describe squares of DF, AF: therefore, the two circles, intersecting squares of A B, BC, CD, D A are to- one another in C, and A gether equal to twice the squares of join CA, C B. BF, DF, together with four times the Then, because CA, square of AF. But, because the base C B are each of them BD of the triangle FBD is bisected in E, equal to A B (def. 24.), the squares of BF, DF, taken twice, are they are (ax. 1.) equal to one another, (40.) equal to four times the squares of and the triangle CAB is equilateral. BE, E F: therefore the squares of A B, Therefore, an equilateral triangle has BC, CD, D A are together equal to been described upon the given finite four times the squares of A F, B E, to straight line AB, which was required to gether with four times the square be done. that is, to the squares of AC, BD, PROP. 43. Prob. 2. (Euc. i. 10.) (39. Cor. 2.) together with four times To bisect a given finite straight the square of E F. line AB. Therefore, &c. Upon either side of A B (42.) describe Cor. In a parallelogram, and in a parallelogram only, the points E and F an equilateral triangle: join the vertices or summits C,D, and let ČD cut AB in E.. coincide, because (22.) the diagonals bisect one another : therefore, in a paral- CBD have the three sides of the one Then, because the triangles CAD, lelogram, and in a parallelogram only, equal to the three sides of the other, the squares of the diagonals are toge- each to each, (7.) they are equal in every ther equal to the squares of the four respect, and the angle A Č D is equal sides. to BCD. Therefore, the triangles ACE, B C E, having two sides of the SECTION 7.--Problems. one equal to two sides of the other, Upon a piece of paper for a plane, each to each, and the included angles with a pen,* a ruler, and a pair of com- equal to one another, are equal in every passes, it is evident, that, first, a straight respect (4), and A E'is equal to EB. line may be drawn from any one point Therefore, &c. to any other point; 2ndly, a terminated Cor. By the same construction, a straight line may be produced to any straight line may be drawn, which shall length in a straight line ; 3dly, from the bisect any given straight line at right greater of two straight lines, a part may angles. be cut off equal to the less; and 4thly, a N.B. It is evident, from 6. Cor. 4., circle may be described from any centre, that the same end will be obtained by and with any distance from that centre. joining the point of intersection C of any two equal circles described from the Although the paper be not an exact plane, nor centres A, B, and cutting one another the pen such as may serve to draw an exact line, these defects admit of being removed to any required section D of any other two equal circles above A B, with the point of interdegree, and do not, in the least, affect the accuracy of our conclusions, with regard to exact lines and planes described from the same centres, and it is the common section of two planes, see Book IV.? low ARB: for C, D will be the vertices of An edge which is a right line, or nearly so, (because cutting one another either above or bemay be obtained by doubling over a piece of paper upon itself, and a rightangle (def. 10.) by doubling isosceles triangles upon the same base. over this edge upon itself. These are both useful upon occasion, especially the right angle, which is This observation may be of use when so frequently required in geometrical constructions one of the triangles is already described, that a case of instruments is commonly provided as in the first method of Prob. 4. with one. Among practical mechanics it is known under the name of the square. Parallel lines also occur so frequently, that it is convenient to have a PROP. 44. Prob. 3. (Euc. i. 11.) ruler expressly for drawing them, called a parallel To draw a straight line at right ruler. This may be made of two or three different forms, the best of which is that of a flat ruler running angles to a given straight line A B, upon two equal rollers. from a given point .C in the same.; A B B D a E D A CB First method. In CA take any from the centre C, point A, and make CB with the radius. equal to CA: upon CD, describe a A B describe (42.) the circle cutting AB equilateral triangle in the points A, DAB, and join D C. B: bisect (43.) AB Then, because the triangles DCA, in E, and join C E. D C B have the three sides of the one Then, because the triangles CEA, equal to the three sides of the other, CEB have the three sides of the one each to each, (7.) the angle DCA is equal to the three sides of the other, equal to DCB; and they are adjacent each to each, (7.) the angle C E A is angles; therefore each of them is a right equal to CEB, and (def. 10.) C E is at angle (def. 10.), and D C is at right an right angles to A B. gles to A B. Second method. Draw from C to Second method. Take any point D A B any straight line CA: bisect CA which is not in AE, and from the radius DA or DC, describe a circle, (43.) in D: from the centre D, with the centre D, with the radius D C describe the circle ACE, cutting cutting A B in a second point E, and join CE. Then, for the same reason, the line AB a second time as in the second method in the point A : join AD, of the preceding, proand produce it to meet the blem, CE is at right ancircumference in E, and gles to A B. Also, if join E C. Then because the points A, E coinin the triangle E CA, the cide, CA is at right line D C drawn from the angle C to the middle of the opposite side equal to angles to A B. Third method. In A B take any two half that side, the angle C is (19. Cor. 4.) a right angle, and C E is at right angles with the radii A C, BC, describe circles points A, B, and from A, B, as centres, to Å B. If the points A, C, coincide, cutting one another a DC is (12. Cor. 3.) at right angles to second time in F: join AB. CF, and let CF cut Third method. Take any straight A B in E. Then, beline M, and make C A equal to four cause ACF, BCF, times M: from the centre C, with a are isosceles triangles radius equal to three times M, describe upon the same base a circle : from the centre A, with a CF, the line AB which radius equal to five joins their summits, bisects the base at times M, describe a right angles (6. Cor.4.); that is, C E is second circle, cutting at right angles to AB. the former in D, and Therefore, &c. join DC, D A. Then, When C A is equal to CB, (which because the square of five times M is (29. Cor. 3.) equal to may happen, when the points A, B are on different sides of E,) the last of these the square of four times M, together three methods is, in practice, nearly the with the square of three times M, the same as the first. The last two methods square of A D is equal to the squares are applicable to the case, in which the of AC, CD; and, therefore, (36.) the point Cfalls near the edge of the paper. angle A C D is a right angle, or D C is at right angles to A D. PROP. 46. Prob. 5. (Euc. i. 9.) Therefore, &c. To bisect a given rectilineal angle BAC. The last two methods are particularly In A B take any point B : make AC convenient, when the given point C is equal to AB: join BC: upon B C denear the edge of the paper. scribe (42.) the equilateral triangle BD C, and join Prop. 45. Prob. 4. (Euc. i. 12.) A D. Then, because the tri angles A B D, AC D have To draw a straight line at right angles the three sides of the one to a given struight line A B, from a equal to the three sides of given point C without it. the other, each to each, (7.) First method. Take any point D upon the angle B A D is equal to the other side of A B, and join CD: the angle CAD. A B 1C C B M a B Therefore, &c. centre B with a radius equal to D E, Cor. By repeating this process with describe a circle cutting the last circle in the halves, quarters, &c. of the given F; and join FA, FB. Then, because angle, it may be divided into four, eight, the triangles ECD, FAB have the &c., equal parts. three sides of the one equal to the three Scholium. sides of the other, each to each, (7.) they There is no construction in Plane at A is equal to the angle at C. are equal in every respect, and the angle Geometry, i. e. no construction practi- Therefore, &c. cable with the assistance of the right line and the circle only, by which a given PROP. 48. Prob. 7. (Euc. i. 31.) angle may be divided into three, five, Through a given point A, to draw a &c. equal parts. It is true that, by the straight line parallel to a given straight description of other lines which are line B C. plane curves, (as the hyperbola, con- From A to BC choïd, and cissoïd,) any given angle draw any straight may be trisected, or divided into three line A Č, and at equal parts ; but such constructions, the point A make not being effected by the two species of the angle CAD lines which are the subject of these Ele- equal to ACB ments, are for certain reasons (to be (as directed in Prop. 47., or as in the found in the application of Algebra to adjoined figure) : then, because these Geometry) said to be out of the range angles are alternate angles, A D is (15. of Plane Geometry. There are, how. Cor. 2.) parallel to B C. ever, some particular cases in which, by Therefore, &c. means of the circle, an angle may be N.B.—In the figure, equal circles are, divided into three, five, and other num- in the first place, described from the bers of equal parts. Such is the divi- centres A, C, the former cutting BC a sion of a right angle into three equal parts. second time in B; and the point D is In this case we are already aware, that, then determined by describing a circle since an equilateral triangle has its three from the centre A with the radius B C. angles equal to one another, and the three This construction is equally short, in taken together equal to two right angles, practice, with that of Prop.47.; and when each of them is two thirds of a right angle. it is required to obtain the point D at as The problem, therefore, will be solved great a distance as possible from A, may by describing an equilateral triangle be preferred, the line AC being drawn upon either of the legs, taken of any more obliquely for that purpose. length at pleasure. In like manner, by means of an isosceles triangle, each of PROP. 49. Prob. 8. (Euc. vi. 9). whose base-angles is double of the verti- To divide a given straight line, A B, cal angle, (see book iii.) and is, there- into any number of equal parts. fore, four-fifths of a right angle, Plane Let it be required to divide A B into Geometry enables us to divide a right five equal parts. Draw A C, making any angle into five equal parts. But these angle with A B, and taking any distance are only particular cases, and indicate M, set it off upon A C five times from no general process. A to C: join C B, and through the points С of division in A C, viz, D, E, &c. draw PROP. 47. Prob. 6. (Euc. i. 23.) (48.) parallels to C B, cutting A B in the At a given point A, in a given points F, G, &c. AB shall be divided straight line AB, to make an angle at the latter points into five equal parts, equal to the given rectilineal angle č. Through F and From the centre C, with any radius, G draw (48.) F describe a circle, cutting the sides of H and GK, the given angle in the points D E, and each of them parallel to AC. hen, because FC is a parallelogram, FH is equal to D C (22.); and for the like reason GK is equal to join D E: from the centre A, with a ED: but ED, DC, are equal to one radius equal to C E or C D, describe a another; therefore (ax. 1.), FH is equal circle, cutting A B in B, and from the to G K. But FH, GK, are sides of the a D X H А C: B e B |