3, 4, &c., will be effected by dividing the numerical value by 2, 3, 4, &c., as the case may be. 3. But for a ready and infallible method of bisecting a line, Prop. 10 cannot be dispensed with. PROP. 11.-PROB. To draw a st. line at right angles to a given st. line from a given point in the same. SOL.-P. 3. From the greater of two st. lines to cut off a part equal to the less. P. 1. On a given st. line to construct an equilateral triangle, Pst. 1. Two points may be joined by a line. DEM.-P. 8. In two triangles, if two sides and the base of one triangle be equal to the two sides and the base of another triangle, the angle between the two equal sides of the one is equal to the angle between the two equal sides of the other. Def. 10. When a st. line on another st. line makes the adjacent angles equal, each of the angles is a right angle. Ax. 1. Things equal to the same thing are equal to one another. make CE = CD; 2 P.1&Pst.1 on DE construct the equil. ▲ FDE, and join 3 Sol. FC. then SDCF and ECF are equal, and consequently rt. angles. DEM. 1 by C. 1 & 2... DC = EC, FC common to the As DCF, 234 L 4 5 Recap. Def. 10... the LS DCF and ECF are rt. angles. Q.E.F. be shown that two st. lines cannot have a common EXP. 1 If possible, let the segment AB be common to the st. lines ABC and ABD. CONS.1 by P. 11. . At. B draw BE, DEM. 1 by Hyp. 2 Def. 10. 3 Hyp. 4 Def. 10. st. lines AB and BC make one st. line ABC, and ABE is a rt. angle; .. LABE = LEBC: Again, st. lines AB and BD form one st. line ABD, and ABE is a rt. angle; 5 D. 2, 4, Ax. 1 Wherefore, 6 Recap. impossible. LEBD: EBD = EBC; which is Therefore, two st. lines cannot have a common SCH.-1. This corollary should be ranked among the axioms; for it is assumed in Prop. 4, where it is taken for granted that if certain st. lines, placed on one another, coincide for any portion of their length, they must coincide throughout. It is also assumed in Prop. 8. 2. When the given point is at the extremity of a given st. line, the line from that extremity should be produced, and the rt. angle be then constructed: 3. Or the following method may be adopted-Let AB be the given st. line, and. B the extremity from which the perpendicular is to be raised. Take any point G above the line AB, and with radius GB describe a circle cutting AB in . H; join HG and produce it to M; or from . H set the same radius GB on HK, KL, and LM; M is the point from which if a st. line be drawn A to B, that st. line MB will be perpen dicular to AB. The radius equals the chord of 60°; three times 60° equal L 180°; and 180° is the semicircle: and by P. 31, bk. iii., the angle HBM in a semicircle is a rt. angle. USE AND APP.-1. By this proposition the Square is constructed, an instrument employed for ascertaining the perpendicular to an horizontal line, and for all purposes for which right angles are needed. 2. On a given st. line AB, to describe an isosceles triangle of which the perpendicular height CD, is equal to the base AB. CONS. 1 by P. 10, 11, & 3 Bisect AB in. D, and make st. line DC perpen. 2 Pst. 1. DEM. 1 by C. 1. 2 P. 4. 3 C. 1. and equal to AB; join AC and BC; the figure ABC is the isosc. A required. in as ADC, BDC, AD=DB, DC is common, and ADC 4 BDC; .. AC=BC. And the perp. DC has been made equal to AB. Q.E.F. PROP. 12.-PROB. To draw a perpendicular to a given st. line of unlimited length from a given point without it. SOL.-Pst. 3. A circle may be drawn from any centre at any distance from that centre. P. 10. To bisect a given st. line. Pst. 1. A line may be drawn from one point to another. DEM.-Def. 15. The radii of the same circle are all equal. P. 8. If two triangles have two sides in one equal to two sides in the other, and the base equal to the base, the angles contained by the two pair of equal sides are equal. Def. 10. A st. line at rt. augles to another st. line, is perpendicular to it. F EXP. 1 Data. 2 Quæs. CONS.1 by Assum. 2 Pst. 3. K Given the st. line AB, and the. C out of it; required from. C a perpendicular to AB. Take any. D on the other side of AB; from. C with rad. CD draw the arc EGF, cutting AB in. F and . G; 3 P. 10, Pst. 1. bisect FG in H, and join C and H, C and F, and C and G; 4 Sol. DEM. 1 byC.3,Def15 2 P. 8. 3 C. Remk. 4 Def. 10. then the st. line CH is perpendicular to AB. in As FHC, GHC, FH = HG, CF = CG, and HC common, Now these are adjacent angles; .. the st. line CH is perpendicular to AB. Therefore, from the given point, &c. Q.E.F. SCH.-1. The properties of the circle form the subject of the third book, but in the construction of the 12th Prop., the Lemma is borrowed from Prop. 2. bk. iii., that the circle will intersect the line in two points. In the triangles NCO, NMO,-the three sides of the one are equal to the three sides of the other, and by P. 8, the angle CNO equals the angle M NO: thus in the triangles NDC, NMD, two sides and the included angle of one are equal to two sides and the included angle of the other :--therefore, by P. 4, ang. ADC equals ang. ADM, and they are adjacent angles; hence st. line CD is perpendicular to st. line AB. USE AND APP.-1. In practice, the problem will be solved by drawing, as in the figure to P. 12, the arc FDE, and from the points F and G, with equal radii, describing arcs intersecting in . K; by joining CK, the perpendicular to AB will be drawn. 2. This problem is indispensable to all Artificers, Surveyors, and Engineers. PROP. 13.-THEOR. The angles which one st. line makes with another upon one side of it are either right angles, or together equal to two right angles. CONS.-P. 11. To draw a st. line at right angles to a given st. line from a given point in the same. DEM.-Def. 10. When a st. line standing on another st. line makes the adjacent angles equal, each of the angles is a rt. angle. Ax. 8. Magnitudes which exactly fill the same space are equal. Ax. 1. Magnitudes which are equal to the same, are equal to each EXP. 1 by Hyp. Let the st. line AB make angles with 2 Concl. 1. then the Ls CBA, ABD, are two rt. A B D CASE I.-Suppose that the ang. CBA is Def. 10. then each of the angles |