.. Then BC is a transversal to the parallels AB and CE; LABC=LBCE. Also, AC is a transversal to the same parallels ; (74°, 1) Cor. An external angle of any triangle is equal to the sum of the opposite internal angles. (49°, 3) 77°. From the property that the sum of the three angles of any triangle is a straight angle, and therefore constant, we deduce the following— I. When two angles of a triangle are given the third is given also; so that the giving of the third furnishes no new information. 2. As two parts of a triangle are not sufficient to determine it, a triangle is not determined by its three angles, and hence one side, at least, must be given (66°, 1). 3. The magnitude of any particular angle of a triangle does not depend upon the size of the triangle, but upon the form only, i.e., upon the relations amongst the sides. 4. Two triangles may have their angles respectively equal and not be congruent. But such triangles have the same form and are said to be similar. 5. A triangle can have but one obtuse angle; it is then called an obtuse-angled triangle. A triangle can have but one right angle, when it is called a right-angled triangle. All other triangles are called acute-angled triangles, and have three acute angles. 6. The acute angles in a right-angled triangle are complementary to one another. 78°. Theorem.—If a line cuts a given line it cuts every parallel to the given line. N Р M Let L cut M, and let N be any parallel to M. Then L cuts N. Proof. If L does not cut N it is || to N. But M is to N. Therefore through the same point P two lines L and M pass which are both || to N. But this is impossible; And N is any line || to M. L cuts N. L cuts every line || to M. (70°, Ax.) q.e.d. 79°. Theorem.—If a transversal to two lines makes the sum of a pair of interadjacent angles less than a straight angle, the two lines meet upon that side .. G of the transversal upon which these interadjacent angles lie. A B F C D and /H GH is a transversal to AB and CD, LBEF+LEFD < 1. Then AB and CD meet towards B and D. Proof. Let LK pass through E making _KEF=LEFC. Then LK is || to CD. it cuts CD. But AB cuts LK in E, (78°) Again, EB lies between the parallels, and AE does not, the point where AB meets CD must be on the side BD of the transversal. g.e.d. Cor. Two lines, which are respectively perpendicular to two intersecting lines, intersect at some finite point. D BL 80°. Def.-1. A closed figure having four lines as sides is in general called a quadrangle or quadrilateral. Thus ABCD is a quadrangle. 2. The line-segments AC and BD which join opposite vertices are the diagonals of the quadrangle. 3. The quadrangle formed when two parallel lines intersect two other parallel lines is a parallelogram, and is usually denoted by the symbol J. 81°. Theorem.-In any parallelogram 1. The opposite sides are equal to one another. 2. The opposite internal angles are equal to one another. 3. The diagonals bisect one another. AB is || to CD, and AC is || to BD, and AD and BC are diagonals. 1. Then ABCD and AC=BD). B Proof.. AD is a transversal to the parallels AB and CD, and ·.· AD is a transversal to the parallels AC and BD, (74°, 1) (74°, 1) (59°) q.e.d. 2. LCAB=LBDC and LACD=LDBA. Proof. It is shown in I that LCAD=LADB and LBAD 82°. Def. 1.-A parallelogram which has two adjacent sides equal is a rhombus. Cor. 1. Since AB=BC (hyp.) B =DC (81°, 1)=AD. A Therefore a rhombus has all its sides equal to one another. Cor. 2. Since AC is the right bisector of BD, and BD the right bisector of AC, C D (54°) Therefore the diagonals of a rhombus bisect one another at right angles. Def. 2.-A parallelogram which has one right angle is a rectangle, and is denoted by the symbol . Cor. 3. Since the opposite angle is a and the adjacent angle is a Therefore a rectangle has all its angles right angles. (81°, 2) (74°, 3) Cor. 4. The diagonals of a rectangle are equal to one another. Def. 3.—A rectangle with two adjacent sides equal is a square, denoted by the symbol ☐. Cor. 5. Since the square is a particular form of the rhombus and a particular form of the rectangle, Therefore all the sides of a square are equal to one another; all the angles of a square are right angles; and the diagonals of a square are equal, and bisect each other at right angles. 84°. Theorem.-If three parallel lines intercept equal segments upon any one transversal they do so upon every transversal. AE is a transversal to the three parallels AB, CD, and EF, so that AC=CE, and BF is any other transversal. Then BI) =DF. Proof. Let GDH passing through D be to AE. Def.-The figure ABFE is a trapezoid. Therefore a trapezoid is a quadrangle having only two sides parallel. The parallel sides are the major and minor bases of the figure. Or, the line-segment joining the middle points of the nonparallel sides of a trapezoid is equal to one-half the sum of the parallel sides. Cor. 2. When the transversals meet upon one of the extreme parallels, the figure AEF' becomes a ▲ and CD' becomes a line passing through the middle points of the sides AE and AF', and parallel to the base EF'. E D Therefore, I, the line through the middle point of one side of a triangle, parallel to a second side, bisects the third side. And, 2, the line through the middle points of two sides of a triangle is parallel to the third side. 85°. Theorem.—The three medians of a triangle pass through a common point. CF and AD are medians intersecting in O. B Proof. Let BO cut AC in E, and let AG || to FC meet BO in G. Join CG. Then, BAG is a ▲ and FO passes through the middle of AB and is || to AG, F (84°, Cor 2) .. O is the middle of BG. |