E C 7/4 D A 4o Make the angles A and B each equal to half a right angle; C being a rt. angle, the square on AC will be one half of the square on AB: again, at A and C make the angles CAD, ACD each equal to half a rt. angle, and the square on CD will be of that on AC, or of the square on AB: the process may be continued for any bisection of the rt. angles supposed to be formed at the extremities of a line, as for 1, 16, 32, 1 &c. B COR. 4.-"If a perpendicular BD be drawn from the vertex of a triangle to the base, the difference of the squares of the sides A B and CB, is equal to the difference between the squares of the segments AD and DC." For (47. I.) AB2 = AD2 + BD2; and CB2 DC2 + BD2. Take the latter from the former, B B D CA C D 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." 2 = For AB2+ BC2 A B2 + BD2 + DC2; and A B2 + BC2 BC2 + AD2 + BD2; therefore (Ax. 1) A B2 + BD2 + DC2 = BC2 + AD2 + BD2. Take away the common square BD2; and (Ax. 3) AB2 + CD2 BC2 + AD2. SCHOLIUM.-1. The 32nd and 47th Propositions are said to have been discovered by PYTHAGORAS, born B. C. 570, and other principles of Geometry to have been brought by him from Egypt into Greece. Whatever we may think of the tale of his extravagant joy on the discovery of the 47th, certain it is that this is one of the most important propositions in all Euclid, for on it, and on the proposition in the sixth book which establishes the similitude of equiangular triangles, the whole science of Trigonometry is founded. 2. A plain and practical illustration of the 47th Proposition 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 AB4, into sixteen, and that on BC 5, into twenty-five squares: the sum of the squares, 9+16, on AC and AB, equals the squares on BC. To form a rt. angle, lines containing 3, 4, and 5 equal parts, or any equimultiples of them, may be used: thus, measure off a line containing 5 feet or links, &c.; at one extremity B, with 4 feet or links, draw an arc, and at the other extremity C, with 3 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 right angles to each other. USE AND APPLICATION.-1. The 47th and its corollaries may be applied for the construction of all similar rectilineal figures by Prop. 31, bk. vi., where it is proved "that in right-angled triangles, the rectilineal figure described upon the side opposite to the right angle, is equal to the similar and similarly-described figures upon the sides containing the right 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." 66 1o. To make a rectilineal figure ADKEF, similar to a given rectilineal figure ABLHC. C 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, ÉF parallel to HC. Then ADKEF will be similar to ABLHC,-for by F/ 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 ALH, and AEF to AHC. Now, by Prop. 19, bk. vi., "Similar triangles are to each other as the squares of their like sides;" and figures made up of similar triangles being similar, all similar figures are to each other as the squares of their like sides. area AEF AE2 AK2 For, area AHC In like manner = = AH2 AL2 area AKE AD2 area AHC AB2 = AD2 area AEF AD2 area ALH AD2 A B2 E 2o. To make a circle the double or the half of another circle. Let A B 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 on BE; and the circle of which AB is the diameter onehalf of the circle of which B E is the diameter. A D 2. By this 47th Proposition, the Chords, Natural Sines, Tangents, and Secants of Trigonometrical Tables are constructed. With AC as radius desc. G an arc CG, and from A raise Let it be supposed that the Radius AB is divided into 1,000,000 parts, and that the arc BC is 30°. Since the Chord Ck of 60° is equal to the Radius AC; BD the sine of 30° shall be equal to the half of AC, or 500000, in the rt. angled triangle ADB. Now AB2 = AD2 + BD2; and AB2-BD2 = AD2 or BF2, the Sine of the complement: substituting the numbers we have √ (10000002—5000002) 866025 = FB. Next, as the triangles ABD, AEC are equi-angular, we have the proportion AD: BD AC: CE; therefore, the tangent of 30° CE= 500000 × 1000000 or, 866025 50000000000 = BD. AC, = 577350. Then AC2 + CE2 = AE2, and AE is the secant of 30°;-therefore √ (10000002 + 5773502) = 1154703 the Natural Secant for an arc of 30°. 3. Arithmetically, when the given multiples, the sum of the squares of the greater, as (6 × 6) + (8 × 8): = 10 × 10 numbers are 3, 4, 5, or their equitwo less is equal to the square of the 100, and when any two are given we can find the third exactly: but with respect to all other numbers, though the sum of the squares of any two numbers always equals or constitutes the square of a number greater than either, we cannot attain that number with perfect accuracy; excepting in the case of right triangular numbers, all we can do is to approach its value by increasing the number of decimal places in the root. Thus (5 x 5) + (8 × 8) = 25+ 64 89; but the square root of 89 is 9433981, &c.-See the Introduction, § vi., on Incommensurable Quantities. = 4. 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 + the 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 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. = 5. 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 VAB2 + 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? AB the horizontal radius. Here CBCL + LB. And √CB2—AC2. 126 Ex. 2. A meteor B is seen over a distance from A to D of 200 miles: required its height. 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 L or A? Here BC-LC BL. And √(40002 +12)= 4000.0001255 = BC: then 4000.001255-4000.0001255 of a mile = 8 inches nearly. 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, A B2 BL x 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. 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. CONSTRUCTION.-P. 11. To draw a st. line at rt. angles to a given st. line from a given point in it. P. 3. From the greater to cut off a part equal to the less. DEMONSTRATION.-Ax. 2. If equals be added to equals, the wholes are equal. EXP. P. 47. In a rt. angled triangle the square on the side subtending the rt. angle is equal to the sum of the squares of the sides containing the rt. angle. Ax. 1. Magnitudes equal to the same magnitude are equal to each other. P. 8. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them of the other. 1| Hyp. 2 Concl. CONS. 1 P. 11. AC; B 2P.3 & Pst.1 make AD AB, and join DC. = D |