PROPOSITION I. PROBLEM. To construct a triangle, of which the three sides are given. Let AB represent the base, and G, H two sides of the . triangle which it is required to construct. From the centre A, with the distance G, describe a circle; and, from the centre B, with the distance H, describe another circle, meeting the former in the point C: ACB §: same circle are equal, AC is equal to G ; and, for the same reason, BC is equal to H. Consequently the triangle ACB answers the conditions of the problem. The limiting circles, after mutually intersecting, must obviously diverge from each other, till, crossing the extension of the base AB, they return again and meet below it; thus marking two positions for the required triangle. Corollary. If the radii G and H be equal to each other, the triangle will evidently be isosceles; and if those lines be likewise equal to the base AB, the triangle must be equilateral. PROP. II. THEOREM. Two triangles are equal, which have all the sides of the one equal to those of the other. Let the two triangles ABC and DFE have the side AB equal to DF, AC to DE, and BC to FE: These triangles are equal. ~ * For conceive the triangle ACB to be applied to DEF: The point A being laid on D, and the side AC on DE, their other extremities C and E must coincide, since AC is equal to DE. And because AB is equal to DF, the point B must occur in the circum- ference of a circle described from `- B D with the distance DF; and, for the same reason, B must be found in the circumference of a circle de D IE A. scribed from E with the distance EF: The vertex of the triangle ACB must, therefore, appear in a point which is common to both those circles, or, by the first proposition, in F the vertex of the triangle DFE. Consequently these two triangles, being rectilineal, must entirely coincide. The angle CAB is equal to EDF, ACB to DEF, and CBA to EFD ; the equal angles being thus always opposite to the equal sides. Scholium. This proposition is only the preceding one changed into a theorem. But any rectilineal figure may be divided into triangles, which, being separately constructed with the same corresponding sides, must, by their combination, hence form an equal figure. PROP. III. THEOR. Two triangles are equal, if two sides and the angle contained by these in the one be respectively equal to two sides and the contained angle in the other. Let ABC and DEF be two triangles, of which the side AB is equal to DE, the side BC to EF, and the angle ABC contained by the former equal to DEF which is contained by the latter: These triangles are equal. For let the triangle ABC be applied to DEF: The vertex B being placed on E, and the side BA on ED, the extremity A must fall upon D, since AB is equal to DE. And because the angle or divergence ABC is equal to DEF, and the side AB coincides with DE, the other side BC must lie in the same direction with EF, and being of the same length, must terminate with it; and consequently, the points A and C resting on D and F, the straight lines AC and DF will also coincide. Wherefore, the one triangle being thus perfectly adapted to the other, a general equality must obtain between them : The third sides AC and DF are hence equal, and the angles BAC, BCA opposite to BC and BA are equal respectively to EDF and EFD, which the corresponding sides EF and ED subtend. Schol. By applying this proposition to practice, the mutual distance may be found between two remote objects which have their communication obstructed. PROP. IV. PROB. At a point in a straight line, to make an angle equal to a given angle. At the point D in the given straight line DE, to form an angle equal to the given angle BAC. In the sides AB and AC of the given angle, assume the points G and H, join GH, from DE cut off DI equal A. ID to AG, and on DI consti tute (I. 1.) a triangle DKI, H. K having the sides DK and G IA F IK equal to AH and GH: * IE EDK or EDF is the angle required. For all the sides of the triangles GAH and IDK being respectively equal, the angles opposite to the equal sides must be likewise equal (I. 2.), and consequently IDK is equal to GAH. Cor. If the segments AG, AH be taken equal, the construction will be rendered simpler and more commodious. Schol. By the successive application of this problem an angle may be continually multiplied. Two circles CEG. and ADF being described from the vertex B of the given angle with radii BC and BA equal to its sides, and the base AC being repeatedly inserted between those circumferences; a multitude of triangles will be thus formed, all of them equal to the original triangle ABC. Consequently the angle ABD is double of ABC, ABE triple, ABF quadruple, ABG quintuple, &c. If the sides AB and BC of the given angle be supposed equal, only one circle would be required, a series of equal isosceles triangles being constituted about its centre. It is evident that this addition is without limit, and that the angle so produced may continue to spread out, and its opening side even make repeated revolutions. PROP. V. PROB. To bisect a given angle. Let ABC be an angle which it is required to bisect. In the side AB take any point D, and from BC cut off BE equal to BD ; join DE, on which construct (I. 1.) the isosceles triangle DFE, and draw the straight line BF: The angle ABC is bisected by BF. For the two triangles DBF and EBF, having the side DB equal to EB, the side DF to EF, and BF common to both, are (I. 2.) equal, and consequently the angle DBF is equal to EBF. Cor. Hence the mode of drawing a perpendicular from a given point B in the straight line AC; for the angle ABC, which the opposite segments BA and BC make with each other, being equal to two right angles, the straight line that bisects it must be the per pendicular required. Taking -j-H i.T.C BD, therefore, equal to BE, and |