CONSTRUCTION.-From the point A draw AC at right Therefore AB is equal to DE, and AD to BE (I. 34). Therefore the four straight lines BA, AD, DE, EB are equal to one another (Ax. 1), And the parallelogram ADEB is therefore equilateral. For since the straight line AD meets the parallels AB, DE, the angles BAD, ADE are together equal to two right angles (I. 29). But BAD is a right angle (Const.); Therefore also ADE is a right angle (Ax. 3). But the opposite angles of parallelograms are equal (I. 34); angle (Ax. 1); It is equilateral. tangular. Therefore the figure ADEB is rectangular; and it has It is recbeen proved to be equilateral; therefore it is a square (Def. 30). Therefore, the figure ADEB is a square, and it is described ..a square. upon the given straight line AB. Q. E. F COROLLARY.-Hence every parallelogram that has one right angle has all its angles right angles. Proposition 47.-Theorem. In any right-angled triangle, the square which is described upon the side opposite to 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 shall be equal to the -- squares described upon BA, AC. CONSTRUCTION. On BC describe the square BDEC (I. 46). On BA, AC describe the squares GB, HC (I. 46). Through A draw AL parallel to BD or CE (I. 31). Join AD, FC. PROOF.-Because the angle BAC is a right angle (Hyp.), and that the angle BAG is also a right angle (Def. 30), The two straight lines AC, AG, upon opposite sides of AB, make with it at the point A the adjacent angles equal CG is a to two right angles; straight A ABD == Therefore CA is in the same straight line with AG (I. 14). For the same reason, AB and AH are in the same straight line.' Now the angle DBC is equal to the angle FBA, for each of them is a right angle (Ax. 11); add to each the angle ABC. Therefore the whole angle DBA is equal to the whole angle FBC (Ax. 2). And because the two sides AB, BD are equal to the two sides FB, BC, each to each (Def. 30), and the angle DBA equal to the angle FBC; Therefore the base AD is equal to the base FC, and the triangle ABD to the triangle FBC (I. 4). Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD, and between the same parallels BD, AL (I. 41). And the square GB is double of the triangle FBC, because they are on the same base FB, and between the same parallels parallelo- FB, GC (I. 41). Hence gram BL =square GB, and parallelo gram CL HC. But the doubles of equals are equal (Ax. 6), therefore the parallelogram BL is equal to the square GB. In the same manner, by joining AE, BK, it can be shown quare that the parallelogram CL is equal to the square HC. Therefore the whole square BDEC is equal to the two squares GB, HC (Ax. 2); And the square BDEC is described on the straight line BA2+AC2 BC, and the squares GB, HC upon BA, AC. Therefore the square described upon the side BC is equal to the squares described upon the sides BA, AC. Therefore, in any right-angled triangle, &c. Q. E. D. Proposition 48.-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. Let the square described upon BC, one of the sides of the triangle ABC, be equal to the squares described upon the other sides BA, AC; The angle BAC shall be a right angle. Draw CONSTRUCTION.-From the point A draw AD at right right angles to AC (I. 11). angles to AC. (Do not produce Make AD equal to BA (I. 3), and join DC. is equal to the square on BA. To each of these add the square on AC. Therefore the squares on DA, AC are equal to the squares on BA, AC (Ax. 2). But because the angle DAC is a right angle (Const.), the square on DC is equal to the squares on DA, AC (I. 47), And the square on BC is equal to the B squares on BA, AC (Hyp.); D Then Therefore the square on DC is equal to the square on BC DC2 = (Ax. 1); And therefore the side DC is equal to the side BC. And because the side DA is equal to AB (Const.), and AC common to the two triangles DAC, BAC, the two sides DA, AC are equal to the two sides BA, AC, each to each. And the base DC has been proved equal to the base BC; BC2 Hence DAC= 4 BAC. Therefore the angle DAC is equal to BAC (I. 8). Therefore also BAC is a right angle (Ax. 1). EXERCISES ON BOOK I. 1. From the greater of two given straight lines to cut off a portion which is three times as long as the less. 2. The line bisecting the vertical angle of an isosceles triangle also bisects the base. 3. Prove Euc. I. 5, by the method of super-position. 4. In the figure to Euc. I. 5, show that the line joining A with the point of intersection of BG and FC, makes equal angles with AB and AC. 5. ABC is an isosceles triangle, whose base is BC, and AD is perpendicular to BC; every point in AD is equally distant from B and C. 6. Show that the sum of the sum and difference of two given straight lines is twice the greater, and that the difference of the sum and difference is twice the less. 7. Prove the same property with regard to angles. 8. Make an angle which shall be three-fourths of a right angle. 9. If, with the extremities of a given line as centres, circles be drawn intersecting in two points, the line joining the points of intersection will be perpendicular to the given line, and will also bisect it. 10. Find a point which is at a given distance from a given point and from a given line. 11. Show that the sum of the angles round a given point are together equal to four right angles. 12. If the exterior angle of a triangle and its adjacent interior angle be bisected, the bisecting lines will be at right angles. 13. If three points, A, B, C, be taken not in the same straight line, and AB and AC be joined and bisected by perpendiculars which meet in D, show that DA, DB, DC are equal to each other. PROP. 16-32. 14. The perpendiculars from the angular points upon the opposite sides of a triangle meet in a point. 15. To construct an isosceles triangle on a given base, the sides being each of them double the given base. 16. Describe an isosceles triangle having a given base, and whose vertical angle is half a right angle. 17. AB is a straight line, C and D are points on the same side of it; find a point E in AB such that the sum of CE and ED shall be a minimum. 18. Having given two sides of a triangle and an angle, construct the triangle. Examine the cases when there will be (1.) one solution; (2.) two solutions; (3.) none. 19. Given an angle of a triangle and the sum and difference of the two sides including the angle, to construct the triangle. 20. Show that each of the angles of an equilateral triangle is twothirds of a right angle, and hence show how to trisect a right angle. 21. If two angles of a triangle be bisected by lines drawn from the angular points to a given point within, then the line bisecting the third angle will pass through the same point. 22. The difference of any two sides of a triangle is less than the third side. 23. If the angles at the base of a right-angled isosceles triangle be bisected, the bisecting line includes an angle which is three halves of a right angle. 24. The sum of the lines drawn from any point within a polygon to the angular points is greater than half the sum of the sides of the polygon. PROP. 33-48. 25. Show that the diagonals of a square bisect each other at right angles, and that the square described upon a semi-diagonal is half the given square. 26. Divide a given line into any number of equal parts, and hence show how to divide a line similarly to a given line. 27. If D and E be respectively the middle points of the sides BC and AC of the triangle ABC, and AD and BE be joined, and intersect in G, show that GD and GE are respectively one-third of AD and BE. 28. The lines drawn to the bisections of the sides of a triangle from the opposite angles meet in a point. 29. Describe a square which is five times a given square. 30. Show that a square, hexagon, and dodecagon will fill up the space round a point. 31. Divide a square into three equal areas, by lines drawn parallel to one of the diagonals. 32. Upon a given straight line construct a regular octagon. 33. Divide a given triangle into equal triangles by lines drawn from one of the angles. 34. If any two angles of a quadrilateral are together equal to two right angles, show that the sum of the other two is two right angles. 35. The area of a trapezium having two parallel sides is equal to half the rectangle contained by the perpendicular distance between the parallel sides of the trapezium, and the sum of the parallel sides. |