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 squares described upon the sides containing the right angle. CON.--P. 46, P. 31, Pst. 1--DEи.--Def. 30, P. 14, Ax. 1, Ax. 2, P. 4, P. 41, Ax. 6. 3 Sim. 4 C.1, Def. 30 5 Ax. 1. 6 Add. 7 Ax. 2. AG on opp. sides of AB make at the A the adj. s two rt. angles, and.. CA is in the same st. line with AG. Also AB is in the same st. line with AH. And. the s DBC and FBA are each a rt. angle); the DBC the FBA: To each of the equals add the ABC the whole DBA = the whole FBC: 8 C.1, Def. 30 Hence, . AB = FB, BD = BC, and included DBA included FBC; AD FC, and A ABDA FBC. BL and the A ABD are both on the same base BD, and between the same ||s BD and AL; BL is double of the ▲ ABD. 9 P. 4. 10 C. 2. Now the 11 P. 41. 12 C. 2. 13 P. 41. 14 D. 9. 15) Ax. 6. 16 C. 3. Sim. 17 Ax. 2. 18 C. 1. 19 Conc. 20 Rec. the Also the square GB and the ▲ FBC are both on the same base FB, and between the same parallels FB and GC; the square GB, is double of the ▲ FBC: But the A ABD= the Δ FBC ; therefore the BL the square GB. CL = the HC; therefore the whole square BDEC = the two squares GB and HC. Now the squares are BE on BC, BG on BA, and CH on Therefore, in any right-angled triangle, &c. COR. 1.- "" If the sides of a rt. angled ▲ be given in numbers, its hypotenuse may be found: for let the s 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 = √ AB2 + AC2. 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. √402+30% or 1600 + 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=√BC2 — AC2; and AC = √BC2 — AB2. 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. 1o. Set the lines AB and BC, representative of the sides of the first two given squares, at rt. angles G D B A E 2o. Supposing AB to represent the line on which the given square is constructed, its multiple 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 AC and AB. 4o Make the angles A and B each half a rt angle, C being a rt. angle the on AC will be one half E D 1/2 A COR. 4.-If a perpendicular BD be drawn from the vertex of a ▲ to the base, the difference of the squares of the sides AB and CB, is equal to the difference between the squares of the segments AD&DC." For (47. I.) AB2 - = AD2+ BD2; and CB2 DC2 + BD2. Take the latter from the former, AB2 B B A A А D CA COR. 5.-" If a perpendicular be drawn from the vertex B to the base AC, or AC produced, the sums of the squares of the sides and alternate segments are equal." For AB2BC2 AB2 + BD2+ DC2; and AB2 + BC2 therefore (Ax. 1) AB2 + BD2 + DC2 = BC2 + AD2 + BD2. 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 AC 3, 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, lines 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 47th and its corollaries 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 circle, by Prop. 2, bk. xii., "that circles are to one another as the squares of their diameters." 1o. To make a rectil. figure ADKEF, similar to a given rectil. figure ABLHC. F Divide the figure into triangles by the lines AH, AL produced, if necessary; take AD equal to the side of the 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 ABLHC,-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. = BC2 + 3 C E H C D B 2o. 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. B C D E APP. 2 The height of any elevation on the earth's surface is so small when compared with the earth's diameter, that 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 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. Ε 3. Heights and distances from the curvature of the earth are computed by Prop. 47, from the principle established in Prop. 16, bk. iii., that the tangent AB is perpendicular to the radius CA of the arc AL. Then if AB be required, we have (LC + LB)2-AC2 = AB: if LB, the formula is AB2 + AC2 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-AC2= AB the horizontal radius. Or, 40022-4000216016004-1600000016004. 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 Or, √40002 + 1002 16010000 AC2 + AB2 = CB 4001.24 miles. And 4001 24-4000 124 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 farther from the earth's centre than L or A? Here BC-LC: BL. And (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, AB2 — BLX EL nearly. 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. PROP. 48.—THEOR.—(Converse of the 47th.) If the 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, Hyp. 2 Conc. C. 1 by P. 11. 2 P. 3&Pst.1 3 Sol. D. 1 by C. 2. 2 Add. 3 Ax. 2. 4 C. 1. 7 Ax 1. then the squares on DA and AC = the squares on AB and AC: But DAC is a rt. angle; ..the square on DC Also the square on BC the squares on AD and AC; the squares on AB and AC; square on BC, and side DC = side BC: 8 C. 2 & D.7 Thus in As DAC, BAC, AD = AB, DC = BC, and 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. Ir 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 used; 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, extending 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. |