EXERCISES. 1. Show conversely, that if the opposite sides of a quadrilateral are equal, the figure is a parallelogram. 2. If two sides of a quadrilateral are equal A and parallel, the figure is a parallelogram. 3. If one angle of a parallelogram is a right angle, the figure is a rectangle. 4. If two parallels are cut by a third straight line, the bisectors of the four interior angles form a rectangle. (See 57, Ex. 3.) 5. If CE is drawn parallel to BD, meeting AD produced, show that BCED is a parallelogram and equal to the parallelogram ABCD. B C D PROPOSITION XXI. THEOREM. E 109. The diagonals of a parallelogram bisect each other. Let the figure ABCE be a parallelogram, and let the diagonals AC and BE cut each other at O. To prove that AO OC and BO = OE. = In the triangles BOC and AOE, BC= AE (by 106), BCO ZOAE, and Z OBC = ≤ OEA (by 62); the triangles are therefore equal (by 88) in all their parts, = or = BO OE and AO = OC. EXERCISES. Q.E.D. 1. Show conversely, if the diagonals of a quadrilateral bisect each other, the figure is a parallelogram, 2. Show that the diagonals of a rhombus bisect each other at right angles. 3. Show that the diagonals of a rectangle are equal. 4. Show that two parallelograms are equal when two adjacent sides and the included angle of the one are equal to the two adjacent sides and the included angle of the other. PROPOSITION XXII. THEOREM. 110. If three or more parallels intercept equal lengths on any transversal, they intercept equal lengths on every transversal. Let AE, BF, CG, and DH be parallels, and MN and OP any two transversals. To prove that if AB = BC = CD, EFFG GH. = then (by 28) EK = FL = GS. M A E B F In the triangles EKF, FLG, and GSH the angles KEF, LFG, and SGH are equal (by 62); also the angles EFK, FGL, and GHS are equal (by 62). or Therefore these triangles are (by 88) equal in all their parts, 111. COR. From the equality of the triangles EKF, FLG, and GSH, KF = LG == SH. HD-DS = SH, but (by 108) CGDS; therefore HD - CG = SH; likewise, CG - BF: LG, and BF— AE = KF. Therefore the intercepted part of each parallel will differ in length from the next intercept by the same amount. PROPOSITION XXIII. THEOREM. 112. The straight line drawn through the middle point of a side of a triangle parallel to the base bisects the remaining side, and is equal to half the base. In the triangle ABC let E be the middle point of AC and DE parallel to BC. A To prove that D is the middle point of AB and that DE = 1⁄2 BC. Through A draw a line parallel to DE, and it will be parallel to BC. B Then AB and AC are transversals cutting parallel lines; therefore (by 110), when AE = EC, AD = DB. Q.E.D. AADE - 0 = DE. 113. The line drawn parallel to the bases through the middle point of one of the non-parallel sides of a trapezoid bisects the opposite side, and is equal to half of the parallel sides. Since DA and CB are transversals, and DC, GH, and AB parallels (by 110), when DG = GA, CH = HB. Q.E.D. Also (by 111) AB – GH = GH – CD, or AB + CD = 2 GH; .. GH = } (AB + CD). PROPOSITION XXV. THEOREM. 114. The three medial lines of a triangle meet in a point which is two-thirds of the way from each angle to the middle of the opposite side. Let ABC be a triangle; P, Q, R, the middle points of its respective sides; BR, AQ, two medial lines of the triangle; O, their point of inter In the triangle AOC, M is the middle point of 40 by construction, and R the middle point of AC by hypothesis; therefore (by 112) RM is parallel to CO and equal to one-half of CO. In the triangle BOC, for the same reason, NQ = CO and is parallel to CO. Therefore (by 28) RM is equal to and parallel with QN. In the triangle ACB (by 112), RQ AB and is parallel to it. = In the triangle AOB (by 112), MN = AB and is parallel to it. Therefore RQ MN and is parallel to it. = Hence the figure RMNQ is (by 99) a parallelogram. Since RMNQ is a parallelogram, (by 109) OR = ON, and MO = OQ. But by construction OM= AM; therefore the three parts AM, MO, and OQ into which AQ is divided are equal. Therefore AO, which contains two of these parts, is twothirds of the whole, or AO AQ, and likewise BO BR. = = Q.E.D. By taking CP and BR as medial lines intersecting at O, and joining S, the middle point of CO, with N, and with R, and drawing RP and NP, it can be shown in the same manner that OC= = CP, and that O is a point on all three medial lines. EXERCISES. 1. The bisectors of the interior angles of a parallelogram form a rectangle. 2. If the non-parallel sides of a trapezoid are equal, the angles which they make with the bases are equal. 3. If from any point in the base of an isosceles triangle parallels to the equal sides are drawn, the perimeter of the parallelogram thus formed is equal to the sum of the equal sides of the triangle. SUGGESTION. See 112. 115. A Polygon is a plane figure bounded by straight lines; as ABCDE. The straight lines are called the Sides of the polygon; and their sum is called the Perimeter of the polygon. The Angles of the polygon are the angles formed by the adjacent sides with each A other; and the vertices of these angles are also called the Vertices of the polygon. 116. The angles of the polygon measured on the side of the enclosed surface are called Interior Angles. B E D An Exterior Angle of a polygon is an angle between any side and the continuation of an adjacent side. A Diagonal is a line joining any two vertices that are not adjacent, as AD. |