49. Scholium. In equal triangles the equal angles are opposite the equal sides. THEOREM XIV.. 50. Two right triangles having the hypothenuse and a side of the one respectively equal to the hypothenuse and a side of the other are equal in all respects. Let A B C have the hypothenuse A B and the side BC equal to the hypothenuse BD and the side BC of B C D ; then are the two triangles equal in all respects. A B C D Place the triangle BCD so that the side BC will coincide with its equal B C, then C D will be in the same straight line with AC (10). An isosceles triangle ABD is thus formed, and BC being perpendicular to the base divides the triangle into the two equal triangles ABC and BCD (43). THEOREM XV. 51. If from a point without a straight line a perpendicular and oblique lines be drawn to this line, 1st. The perpendicular is shorter than any oblique line. 2d. Any two oblique lines equally distant from the perpendicu lar are equal. 3d. Of two oblique lines the more remote is the greater. gle A D E being a right angle is greater than the angle A ED; therefore A D A E (47). 2d. If DE=DC; then the two triangles ADE and ADC, having two sides AD, DE, and the included angle. ADE respectively equal to the two E A sides A D, DC, and the included angle A D C, are equal (40), and AE is equal to A C. 3d. If DB > DE; then, since A ED is an acute angle, AEB is obtuse, and must therefore be greater than ABE (36); hence ABA E (47). 52. Corollary. Two equal oblique lines are equally distant from the perpendicular. THEOREM XVI. 53. If at the middle of a straight line a perpendicular is drawn, 1st. Any point in the perpendicular is equally distant from the extremities of the line. 2d. Any point without the perpendicular is unequally distant from the same extremities. Let CD be the perpendicular at the middle of the line AB; then 1st. Let D be any point in the perpendicular; draw DA and D B. Since C A C B, DA DB (51). 2d. Let E be any point without the perpen dicular; draw E A and EB, and from the point D, where E A cuts D C, draw D B. The an gle ABE > ABD BAD; hence, in the triangle A E B, since the angle ABE > BAD, E A > E B (47). QUADRILATERALS. DEFINITIONS. 54. A Trapezium is a quadilateral which has no two of its sides parallel; as A B C D. A B 55. A Trapezoid is a quadrilateral which has only two of its sides parallel; as EFGH H E F G 56. A Parallelogram is a quadrilateral whose opposite sides are parallel; as I J K L, or M N O P, or Q R S T, or U V W X. 57. A Rectangle is a right-angled parallelogram; as IJ KL. 58. A Square is an equilateral rectangle ; as MNOP. 59. A Rhomboid is an oblique-angled par allelogram; as QRST. 60. A Rhombus is an equilateral rhomboid; as U V W X. X W 61. A Diagonal is a line joining the vertices of two angles not adjacent; as D B. THEOREM XVII. 62. The opposite sides and angles of a parallelogram are equal to each other. Let ABCD be a parallelogram; then A B D C Draw the diagonal B D. As BC and AD are parallel, the alternate angles CBD and BDA are equal (17); and as AB and DC are parallel, the alternate angles A B D and B D C are equal; therefore the two triangles ABD and BD C, having the two angles equal, and the included side B D common, are equal (41); and the sides opposite the equal angles are equal, viz. : A B = DC and BC — A D ; also the angle 4 C, and the angle A ABC ABD+DBC=BDC+BDA=ADC = 63. Cor. 1. The diagonal divides a parallelogram into two equal triangles. 64. Cor. 2. Parallels included between parallels are equal. THEOREM XVIII. 65. If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram. Let ABCD be a quadrilateral having BC equal and parallel to AD; then ABCD is a parallelogram. A B C D Draw the diagonal BD. As BC is parallel to AD, the alternate angles CBD and BDA are equal (17); therefore the two triangles CBD and BDA, having the two sides CB, BD, and the included angle CBD respectively equal to the two sides A D, D B, and the included angle AD B, are equal (40), and DC is equal to A B, and the alternate angles A B D and B D C are equal; therefore A B is parallel to DC (18), and ABCD is a parallelogram. THEOREM XIX. 66. The line joining the middle points of the two sides of a trapezoid which are not parallel is parallel to the two parallel sides, and equal to half their sum. Let EF join the middle points of the sides A B and CD, which are not parallel, of the trapezoid A B C D; then B E G F 1st. EF is parallel to BC and A D. Through F draw G H parallel to B A, meeting AD produced in H. The angles GFC and DFH are equal (11); also the angles GCF and FDH (17) ; and the side CF is equal to FD; therefore the triangles GFC and D F H are equal (41), and GF FIGH A D H But as ABGH is a parallelogram, G II BA (62); therefore FHBA = A E therefore A EFI is a parallelogram (65), and E F is parallel to AD, and therefore also to BC. Now, as the two triangles G F C and D F H are equal, GC DH; therefore, if we add the two equations, we shall have 2 EF AD + BC or EF (AD+BC) |