Scholium.-The angle CDE may be made equal to a right angle or any given angle (I. 13). Proposition 39. Problem. To describe a triangle equal to a given polygon. Let ABCDE be a polygon; it is required to describe a triangle equal to it. Join AC, and draw BF parallel to it, and join AF. Then, because ABC, AFC are on the same base, AC, and between the same parallels, AC, BF, they are equal (I. 35). Similarly, AED, AGD are equal. A Add to both sets of these equals the triangle ACD, and ABCDE is equal to AFG. Corollary. To draw a parallelogram equal to a given polygon. Construct a triangle equal to the polygon, and then a parallelogram equal to the triangle (I. 38). Proposition 40. Theorem.-The complements of a parallelogram are equal to each other. D Let ABCD be a parallelogram, and EH, GF parallelograms about the diagonal; then the complements BK, KD, which make up the whole figure, are equal to each other. Because AC is a diagonal, the triangles ABC, ADC are equal to each other. For the same reason AEK, AHK, and also KGC, KFC, are equal to each other. B A H K F G Hence, AAHK+A KFC=AAEK+A KGC. Subtracting these equals from the whole triangles ABC, ADC, the remainders, BK, KD, will be equal (Ax. 3). Proposition 41. Problem.-To describe a square on a given finite straight line. Let AB be a straight line; it is required to describe a square upon it. From A draw (I. 11) AC at right angles to AB; and make (I. 1) AD equal to AB; through D draw (I. 29) DE parallel to AB; and through B, BE parallel to AD; ADEB is a square. For it is a parallelogram (Const.). Hence AB, DE are equal (I. 32), and AD, BE are D C B E equal; but AB, AD are equal; hence it is equilateral. Also, because AB, DE are parallel, the angles BAD, ADE are together equal (I. 27) to two right angles. But BAD is a right angle (Const.) Hence ADE is a right angle; and because the figure is a parallelogram, the opposite angles are also right angles (I. 32). Therefore the figure is equilateral and rectangular, and is therefore a square. Corollary. If one angle of a parallelogram be a right angle, all the angles are right angles. Proposition 42. Theorem.-In any right-angled triangle the square described on the side opposite the right angle, is equal to the sum of the squares described on the sides containing the right angle. Let ABC be a right-angled triangle, right angled at A; then will, BC2* = AB2 + AC2 *. On BC describe (I. 41) the square BE; and on BA, AC the squares BG, AK. Through A draw AL parallel to BD, Because and join AD, FC. The angle FBA is equal to the angle CBD; to each add ABC; then the angle FBC is equal to the angle ABD. Then, because in the triangles FBC, ABD, FB, BC, and the angle FBC are equal to AB, BD, and the angle ABD, each to each, the triangles are equal (I. 4) in area. B G L E But FBC is half the square BG, because they are on the same base, FB, and between the same parallels, FB, GC. Similarly, ABD is half the parallelogram BL. Therefore the square BG is equal (Ax. 7) to the parallelogram BL. In the same manner it may be proved that AK is equal to CL. Therefore the sum of BG and AK is equal to the whole square BE, or BC2AB+ A C2. Corollary 1.-Hence BC2-AB2-AC2. Corollary 2.-Also, if AB-AC, BC2-2AB=2A C2. Corollary 3.-And if two right-angled triangles have two sides of one equal to the corresponding sides of the other, they will be mutually equilateral and equal in all their parts. * This expression means the square described on BC, etc. EXERCISES ON BOOK I. The proofs or solutions of the following propositions are to be worked out by the student, by the use of those previously demonstrated. In some cases the lines necessary to the construction are drawn, and the propositions which may be needed are referred to. In others, parts of the solution are given. In all cases they are intended to be so clear that a student may, unaided, complete the proof by the exercise of thought and skill. It will often assist him in obtaining a construction to suppose the figure formed, and to note the properties involved; then, by using these properties, he may form the construction. 1. Upon a given straight line to describe an isosceles triangle having each of the equal sides equal to a given straight line. 2. If the angles at the base of an isosceles triangle be bisected, the portions of the bisecting lines between their intersection and the angles bisected, will be equal (I. 14). 3. If a point be taken without a given straight line, the perpendicular is the shortest line that can be drawn from the point to the line (I. 19), (I. 21). 4. The diagonal of a rhombus bisects its angles (I. 9). 5. The four sides of any quadrilateral are together greater than the two diagonals (I. 2). 6. Construct a triangle having given two angles and a side opposite one of them. 7. Why is the problem, to construct a triangle when we have only the three angles, indefinite?-what ambiguity is there when we have two sides and an angle opposite one of them? 8. A straight line parallel to the base of an isosceles triangle makes equal angles with the sides (I. 27). 9. If a line joining two parallels is bisected, show that any line through the point of bisection and terminating in the parallels is also bisected (I. 27), (I. 5). 10. If AB and CD be parallel and FG=EF, show that any other line GH is bisected at K. Draw FL parallel to GH, and work in the triangles FLE, GKF. 11. Trisect a given straight line. G K F A B On the line AB construct an equilateral triangle ABC. B by AD, BD. Bisect the angles at A and Draw DE, DF parallel to AC, CB. Prove that AE= EF = FB. 12. Trisect a right angle. Let ABC be the right angle. On AB construct an equilateral triangle ABD. Bisect the angle ABD by BE. Prove that ABE EBD = DBC (I. 30). = D H Ꮮ E 13. What is the magnitude of an angle of a regular decagon (I. 30 Cor.)? 14. What is the magnitude of an exterior angle of a regular hexagon (I. 30 Cor.)? 15. If the opposite sides of a quadrilateral be equal, it is a parallelogram. 16. If the opposite angles of a quadrilateral be equal, it is a parallelogram. 17. The four triangles into which a parallelogram is divided by its diagonals, are equal in area. |