produced; let them meet in k, and through k draw kl, parallel to ea or fh, and produce ha, gb to the points 1, m: then hlkf is a parallelogram, of which the diameter is hk, and a g, me are the parallelograms about hk; and 1b, bf are the complements: therefore 1b is equal (i. 43) to bf; but bf is equal to the triangle c; wherefore 1b is equal to the triangle c; and because the angle gbe is equal (i. 15) to the angle a bm, and likewise to the angle d; the angle abm is equal to the angle d. Therefore the parallelogram 1b is applied to the straight line a b, is equal to the triangle c, and has the angle abm equal to the angle d. Which was to be done. PROPOSITION XLV.-PROBLEM. · To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. LET abcd be the given rectilineal figure, and e the given rectilineal angle. It is required to describe a parallelogram equal to abcd, and having an angle équal to e. a d f g Join db, and describe (i. 42) the parallelogram fh equal to the triangle adb, and having the angle hkf equal to the angle e; and to the straight line gh apply (i. 44) the parallelogram gm equal to the triangle dbc, having the angle g h m equal to the angle e; and because the angle e is equal to each of the angles fkh, ghm, the angle fk h is equal to ghm: add to each of these the angle khg; therefore the angles fkh, khg, are equal to the angles khg, ghm; but fkh, khgare equal (i. 29) to two right angles; therefore also khg, ghm, are equal to two right angles; and because at the point h in the straight line gh, the two straight b e h lines kh, hm upon the opposite sides of it make the adjacent angles equal to two right angles, kh is in the same straight line (i. 14) with hm; and because the straight line hg meets the parallels km, fg, the alternate angles mhg, hgf are equal (i. 29): add to each of these the angle hg1: therefore the angles mhg, hgl, are equal to the angles hgf, hgl. But the angles mhg, hg1, are equal (i. 29) to two right angles; wherefore also the angles hgf, hg1 are equal to two right angles, and fg is therefore in the same straight line with g1; and because kf is parallel to hg, and hg to ml;kfis parallel (i. 30) to m1; and km, fl are parallels; wherefore kilm is a parallelogram; and because the triangle abd is equal to the parallelogram hf, and the triangle dbc to the parallelogram gm; the whole rectilineal figure abcd is equal to the whole parallelogram kflm; therefore the parallelogram kflm has been described equal to the given rectilineal figure abcd, having the angle fkm equal to the given angle e. Which was to be done. COR. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying (i. 44) to the given straight line a parallelogram equal to the first triangle a bd, and having an angle equal to the given angle. PROPOSITION XLVI.—PROBLEM. To describe a square upon a given straight line. LET ab be the given straight line; it is required to describe a square upon a b. From the point a draw (i. 11) a c at right angles to a b; and make (i. 3) ád equal to a b, and through the point d draw de parallel (i. 31) to a b, and through b draw be parallel to a d therefore a deb is a parallelo. gram: whence a b is equal (i. 34) to de, and a d to be but ba is equal to ad; therefore the C four straight lines ba, ad, de, eb, are equal to one another, and the parallelogram adeb is equilateral, likewise all its angles are right angles; because the straight line ad meeting the parallels ab, de, the angles bad, ade are equal (i. 29) to two right angles: but bad is a right angle; therefore also ade is a right angle; but the opposite angles of parallelograms are equal (i. 34); therefore each of the opposite angles abe, bed is a right angle; wherefore the figure adeb is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line a b. Which was to be done. COR. Hence every parallelogram that has one right angle has all its angles right angles. a PROPOSITION XLVII. THEOREM. In any right-angled triangle, the square which is described upon the side subtending the right angle is equal to the squares described upon the sides which contain the right angle. LET abc be a right-angled triangle having the right angle bac; the square described upon the side bc is equal to the squares described upon ba, a c On be describe (i. 46) the square bdec, and on ba, ac the squares gb, hc; and through a draw (i. 31) al parallel to bd, or ce, and join ad, fc. Then, because each of the angles bac, bag is a right angle (30 def.), the two straight lines ac, ag, upon the opposite sides of a b, make with it at the point a the adjacent angles equal to two right angles; therefore ca is in the same straight line (i. 14) with a g; for the same reason, a b and ah are in the same straight line; and because the angle dbc is equal to the angle fba, each of them being a right angle, add to each the angle abc, and the whole angle dba is equal (2 ax.) to the whole fbc; and because the two sides a b, bd, are equal to the two fb, bc, g f b each to each, and the angle dba equal to the angle fbc; therefore the base ad is equal (i. 4) to the base fc, and the triangle abd to the triangle fbc: now the parallelogram bl is double k(i. 41) of the triangle abd, because they are upon the same base bd, and between the same parallels bd, al; and the square gb is double of the triangle fbc, because these also are upon the same base fb, and between the same parallels fb, gc. But the doubles of equals are equal (6 ax.) to one another therefore the parallelogram bl is equal to the square gb: and, in the same manner, by joining a e, bk, it is demonstrated, that the parallelogram c1 is equal to the square hc; therefore the whole square bdec is equal to the two squares gb, hc; and the square bdec is described upon the straight line bc, and the squares gb, hc upon ba, ac: wherefore the square upon the side bc is equal to the squares upon the sides ba, a c. Therefore, in any rightangled triangle, &c. Q. E. D. d 1 e PROPOSITION XLVIII.-THEOREM. If the square 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 these two sides is a right angle. IF the square described upon bc, one of the sides of the triangle a bc, be equal to the squares upon the other sides ba, a c, the angle bac is a right angle. From the point a draw (i. 11) ad at right angles to a c, and make a d equal to ba, and join dc: Then, because da is equal to a b, the square a of da is equal to the square of a b. To each of these add the square of a c; therefore the squares of d a, a c are equal to the squares of ba, a c. But the square of dc is equal (i. 47) to the squares of da, ac, because dac is a right angle; and the square of bc, by hypothesis, is equal to the squares of ba, ac; therefore the square of dc is equal to the square of bc; and therefore also the side dc is equal to the side bc. And because the side da is equal to cab, and ac common to the two triangles da c, bac, the two da, a c are equal to the two ba, ac; and the base dc is equal to the base bc; therefore the angle dac is equal (i. 8) to the angle bac; but dac is a right angle; therefore also bac is a right angle. Therefore, if the square, &c. Q. E. D. b 333 EXERCISES ON BOOK I. SECT. I.-PROBLEMS. 1. Given a straight line and two points, to find a point in the line equidistant from each of the given points. 2. Given a straight line, a point, and an angle, to draw from the point to the line another straight line that shall make with it an angle equal to the given angle. 3. To divide a given finite straight line into any given number of equal parts. 4. To trisect a right angle, that is, to divide it into three equal parts. 5. Given an angle divided into any number of equal angles, to divide one half of the angle into the same number of equal angles. 6. Given the base, one of the angles at the base, and the sum of the other two sides, to construct the triangle. 7. To find a point in a triangle equidistant from the three angular points of the triangle. 8. Given the three angles of a triangle and the sum of its three sides to construct the triangle. 9. To find a square which shall be equal to any number of given squares. 10. To bisect a parallelogram by a straight line drawn through a given, point in one of its sides. 11. To inscribe a square in a parallelogram. 12. Given a right-angled isosceles triangle to inscribe in it a square.. 13. Given a finite straight line to describe a square of which that line shall be the diameter. 14. To construct a triangle that shall be equal to any given rectilineal figure. D SECT. II.-THEOREMS. 15. There cannot be drawn more than two equal straight lines to another straight line from a given point without it. 16. The straight line drawn from a given point perpendicular to a given line is the shortest that can be drawn from the point to the line. 17. If a triangle have its exterior angle and one of its opposite interior angles double of those in another triangle, its remaining opposite interior angle is also double of the corresponding angle in the other. 18. If two sides of a triangle be bisected, the straight line joining the two points of bisection will be parallel to the other side. 19. The difference between the two sides of a triangle is less than the third side. 20. The sum of two sides of a triangle is greater than twice the line joining the vertex and the middle of the base. 21. Each angle at the base of an isosceles triangle is equal to, or is less, or is greater than the half of the vertical angle, according as the triangle is right, obtuse, or acute-angled. 22. If the vertical angle of an isosceles triangle be bisected, the bisecting line will bisect the base, and be perpendicular to it. 23. In an isosceles triangle, if either of the equal sides be produced beyond the vertex, the line that bisects the exterior angle will be parallel to the base. 24. The diameters of a rectangle are equal to one another. 25. If any number of parallelograms be inscribed in a given parallelogram, the diameters of all the figures shall cut one another in the same point. 26. In a parallelogram the diameters bisect each other. |