BG; C Prop. 47.—THEOR.-(Most Important.) In any right-angled triangle, the square which is described upon the side subtending, or opposite to, the right angle, is equal to the squarcs described upon the sides containing the right angle. CON.--P. 46, P. 31, Pst. 1--DEM.--Def. 30, P. 14, Ax. 1, Ax. 2, P.4, P. 41, Ax. 6. Let ABC be a A,having G 2 Conc. then the square on BCS F Η. AC. K and on AC, a square HC; 2 P. 31. through A draw AL | BD or to CE ; A, E and B, K. are each a rt. angle ; L E AG on opp. sides of AB make at the · A the adj. Ls = two rt. angles, and ... CA is in the same st. line with AG. 3 Sim. Also AB is in the same st. line with AH. 4C.1, Def. 30 And ::: the Zs DBC and FBA are each a rt. angle; 5. Ax. 1. the < DBC = the < FBA : 6 Add. To each of the equals add the < ABC ; 7) Ax. 2. the whole < DBA = the whole _ FBC: 8 C.1, Def. 30 Hence, : AB = FB, BD = BC, and included < DBA included { FBC ; 9) P. 4. AD = FC, and A ABD = A FBC. 10 C. 2. Now the O BL and the A ABD are both on the same base BD, and betwcen the same || BD and AL; 11 P. 41. ... the BL is double of the A ABD. 12 C. 2. Also the square GB and the A FBC are both on the same base FB, and between the same parallels FB and GC; 13) P. 41. .. the square GB, is double of the A FBC : 14 D. 9. But the À ABD = the A FBC ; 151 Ax. 6. therefore the O BL= the square GB. 16 C. 3. Sim. Also, after joining AE, BK, the DCL = the HC ; 17) Ax. 2. therefore the whole square BDEC the two squares GB and HC. 18) C. 1. Now the squares are BE on BC, BG on BA, and CH on 19 Conc. therefore the square on the side BC = the two squares on the sides BA and CA. 20 Rec. Therefore, in any right-angled triangle, &c. CA ; ABP. COR. 1.-" If the sides of a rt. angled A be given 'in numbers, its hypotenuse may be found : for let the Os of the sides be added together, and the square root of their sum will be the hypotenuse.” Suppose AB the base, AC the perpendicular, BC the hypotenuse ; BC EN AB? + AC. Ex. A tower is 40 feet high, its ditch 30 feet broad; required the distance from the farther side of the ditch to the top of the tower. 7402 +30% or V1600 + 900, or 2500 50 ft., distance. COR. 2.-If the hypotenuse and one side be given in numbers, the other side may be found : for let the square of the side be subtracted from that of the hypotenuse, the remainder is equal to the square of the other side. The square root of this remainder will therefore be equal to the other side. Thus, AB = BC? — AC?; and AC=NBC2 COR. 3.- If any number of squares be given, a square equal to their sum may be found; or if 'one square be given, any multiple of it may be ascertained; or if two squares be given, the difference between them ; or a square may be made that shall be the half, fourth, &c., of a given square. 19. Set the lines AB and BC, representative of the sides of the first two given squares, at rt. angles G D ABC; the square on AC = AB2 + BC2 ; at C place CD, to represent the third square, at rt. angles to AC; the square on AD = AB? + BC2 + ČD2 and at D place ED to represent the fourth square, at rt. angles to AD; the square on AE = AB? + BC + CD2 + DEP. 20. Supposing AB to represent the line on which the given square is constructed, its multiple A square will be obtained in a similar way; for in this case BC, CD, DE being each equal to AB, the square on AE is the multiple of the square on AB. 3^. Let AB be the less, and AC the greater of the two lines ; at B the the extremity of the less, raise a perpendicular BG, and from A at the other extremity with AC as radius, inflect on BG the greater line ; the square of the intercept CB = the difference of the squares on A Cand AB. 49 Make the angles A and B each - half a rt angle, C being a rt. angle the on AC will be one half of the on AB : again, at A and C make the s CAD, ACD each equal to half a rt. angle, and the o С. D on CD will be į of that on AC, or of the con AB: Va the process may be continued for any bisection of the A rt. angles supposed to be formed at the extremities of B a line, as for 1-8th, 1-16th, 1-32th, &c. COR. 4.-If a perpendicular BD be drawn from the vertex of a A to the base, the difference of the squares of the B B sides AB and ČB, is equal to the difference between the squares of the segments AD GDC.” For (47. I.) AB? AD+ BDP; and CB2 =DC? + BD?. AB? – CB? - AD? - DC. A D CA CoR. 5.—" If a perpendicular be drawn from the vèrtex B to the base AC, or AC produced, the sums of the squares of the sides and alternate segments are equal.” For AB? + BC = ABP + BD2 + DC2; and AB + BC = BC? + AD? + BDP; therefore (Ax. 1) AB? + BD2 + DC = BCP + AD2 BD? take away the common square BD2 ; and (Ax. 3) AB? + CD? · BC+ AD2 А SCH.--1. A practical illustration of Prop. 47, may be given by taking three lines in the proportion of 3, 4, and 5, and constructing with them a rt, angled triangle BAC; on each of the three sides draw a square, and sub-divide each square; that on AC3, into nine smaller squares; that on AB 4, into sixteen; and that on BC 5, into twenty-five squares; the sum of the squares, 9 and 16, on AC and AB, equals the squares on BC. 2. To form art. angle, .ines containing 3, 4, and 5 equal parts, or any equi-multiples of them, may be used : thus, measure off a line containing 5feet or links, &c.; at one extremity B, with 4 feet or links, draw an arc ; and at the other extremity C, with three feet or links, another arc : the two arcs intersect in A, and the lines from A to C, and from A to B, are at rt. angles to each other. APP.-1. The 47 and its ies may be applied for the construction of all similar rectil. figures by Prop. 1, bk, vi., " that in right angled triangles, the rectil. figure described upon the side opposite to the rt. angle, is equal to the similar and similarly-described figures upon the sides containing the rt. angle;" and to the making of a circle the double or the half of another F circle, by Prop. 2, bk. xii., “that circles are to one another as the squares of their diameters." 1°. To make a rectil. figure ADKEF, similar to a given с rectil. figure ABLHC. Divide the figure into triangles by the lines AH, AL produced, if necessary; take AD equal to the side of the F required figure, and through D draw DK parallel to BL; through K, KE parallel to LH; and through E, EF parallel to HC. Then ADKEF will be similar to ABLAC,- for by Prop. 29, the angles D, K, E, F are equal to those at B, L, H, and C; and the triangles, ADK similar to ABL, AKE to A ALH, and AEF to AHC. н. E B 29. To make a circle the double or the half of another circle. Let AB be the diameter of ADBC; at A raise a perpendicular AE, and at B make the angle ABE equal to half a rt, angle; produce BC until it cuts AE: the square on BE will be double of the square on AB, and the circle of which BE is the diameter double of the circle of which AB is the diameter. Again, let BE be the diameter of a circle; at E and B make angles each equal to half a rt. angle; and the square on AB will be one half of the square BE, and the circle of which AB is the diameter one-half of the circle of which BE is the diameter, D = AB. APP.2 The height of any elevation on the earth's surface is so small when compared with the earth's diameter, that А B for practical purposes, as levelling, and ascertaining the height of mountains, we may consider the earth's actual diameter, and the diameter + elevation, as the same L quantity, i.e., BE and LE not sensibly to differ; nor the arc AL from the horizontal level AB. We assume LE to be 7960 miles, or that we may have an easier number 8000 miles. If we take AB one mile, then BL equals 1-8000th part of a mile, or nearly 8 inches; i.e., for every mile of survey, the surface or curvature of the earth is 8 inches below the horizontal level. 3. Heights and distances from the curvature of the earth are computed by Prop. 47, from the E principle established in Prop. 16, bk. ii., that the tangent AB is perpendicular to the radius CA of the arc AL. Then if AB be required, we have ✓ (LC + LB)2- AC? = AB: if LB, the formula is AB? + AC BC, and BC-LC or AC = BL. Example 1. Given BL the height of the Peak of Teneriffe ; what will be the radius of its horizon, or the distance at which it may be seen ? Here CB=CL + LB. And CB2-AC? = AB the horizontal radius. And V16004 =126 miles Ex. 2. A meteor B is seen over a distance from A to D of 200 miles; required its height. Here BL= BC-LC or AC. And N AC? + AB? = CB And 4001:24-4000 = 1:24 miles height of the meteor. Ex. 3. A fountain B one mile from A, is observed from A to have the same apparent level : how much is B above A ? i.e., how much is B further from the earth's centre than Here BC-LC = BL. And V(40002 + 12) = 4000.0001255 = BC: then 4000.001255-4000 = .0001255 of a mile = 8 inches nearly. 4. By Prop. 36, bk. iii., the square of the tangent AB equals the rectangle of BL into BE : and as in levelling the distances are usually small, AB? BL X EL nearly. When AB is 1 mile, BL is f of 1 foot, or 8 inches ; AB is 2 miles, BL is of 4 feet, or 32 AB is 4 miles, BL is šof 16 feet, or 10'6 feet. Thus two-thirds of the square of the number of miles that the level is long, gives the height of B above A in feet, or what the horizontal level differs from the level of the earth's curvature. or A? If the PROP. 48.—THEOR.—(Converse of the 47th.) squares described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it; the angle contained by the two sides is a rt, angle. Con.-P. 11, P. 3, Pst. 1.- DEM.--Ax. 2, P. 47, Ax. 1, P. 8. E. 1, Нур. D the sum of the squares on AB and AC; with AC; then the < BAC is a rt. L. DA = square on AB : 2 Add. Let the square on À C be added B to each; 3 Ax. 2. then the squares on DA and AC = the squares on AB and AC: 4 C.. But DAC is a rt. angle; 5 P. 47. the square on DC = = the squares on AD and AC; 61 Hyp. Also the square on BC= the squares on AB and AC; 7| Ax l. .. the square on DC = square on BC, and .. side DC = side BC: 8 C. 2 & D.7 Thus in As DAC, BAC, AD = AB, DC = BC, and AC is common ; 9 P.8. _DAC = { BAC: 10 c. 1. But DAC is a rt, angle ; 11 Ax. 1. ... BAC is a right angle. 12 Rec. Therefore if the squares described. &c. Q.E.D. SCH.-The 48th Proposition, may be extended thus :—The vertical angle of a triangle is less than, equal to, or greater than, a rt. angle, as the square on the base is less than, equal to, or greater than, the sum of the squares of the sides. REMARKS ON BOOK I. 1. It will have been seen that the First Book is founded entirely on the Definitions, Postulates, and Axioms ;-the first fixing the meaning of the terms employed; the second assigning the instruments that may be ased ; and the third setting forth the principles on which the comparisons and arguments are conducted. In a few instances for the illustration of certain propositions, other principles, not belonging to the first book, have been assumed ;-but these are to be regarded in their proper light, not as strict proofs, but as methods of explanation. 2. A few only of the properties of the circle are mentioned : those of the straight line and rectilineal angle are subservient to the proof of the properties of the triangle ; and all rectilineal figures are either triangles, or may be resolved into triangles. The First Book therefore, as we have said, may in general terms be described as treating of the Geometry of Plane Triangles. 3. Excluding the Definitions, Postulates, and Axioms, it is not unusual to make a three-fold division of the contents of this Book. The first part, ext ling from the 1st Prop, to the 26th, unfolds the properties of triangles; the second, from Prop. 27 to 32, those of parallel lines ; and the third, from Prop. 33 to 48, those of parallelograms, of course including the square. |