(43.) If straight lines be drawn from the angles of a triangle through any point, either within or without the triangle, to meet the sides, and the lines joining these points of intersection and the sides of the triangle be produced to meet; the three points of concourse will be in the same straight line. Let ABC be a triangle from the three angles of which let lines AF, BE, CD be drawn through a point P within the triangle. Join DE, DF, EF, and produce them H 2 to meet the sides in H, G, I; these three points will be in the same straight line. Join GH, HI. Then the three angles of the triangle DHG being equal to two right angles, as also the three EHI, EIH, and (HEI or) DEF, as also the two DFI, GFI; .. the three angles of the triangle DHG together with the angles EHI, EIH, DEF, DFI, GFI are equal to six right angles. Now the angles of the triangles DEF, FGI are together equal to four right angles, whence DHG, DHI are equal to two right angles; or GH, HI are in the same straight line. SECT. IV. (1.) The diameters of a rhombus bisect each other at right angles. Let ABCD be a rhombus, whose diameters are AC, BD; they bisect each other at right angles in E. Since AB = AD, and AC is common to the two triangles ABC, ADC, the two BA, AC are equal to the two. B E DA, AC, each to each, and BC=DC, .. the angle BAC is equal to the angle DAC. Again, since BA, AE are equal to DA, AE, each to each, and the included angles are equal, .. BE=ED, and the angles AEB, AED are equal, and .. are right angles. For the same reason AE=EC; also the angles BEC, DEC are right angles. (2.) If the opposite sides or opposite angles of a quadrilateral figure be equal, the figure will be a parallelogram. Let ABCD be a quadrilateral figure, whose opposite sides are equal. Join BD. Since AB = DC, and BD is common, the two AB, BD are equal to the two CD, DB, each to each, and AD=BC, .. the angle ABD=BDC, whence (Eucl. i. 27.) AB is parallel to DC; also the angle ADB = DBC, whence AD is parallel to BC; and the figure is a parallelogram. Again, let the opposite angles be equal. Then since the four angles of the quadrilateral figure ABCD are equal to four right angles, and that BAD, ADC together are equal to DCB, CBA, :. BAD, ADC together are equal to two right angles; whence AB is parallel to CD. In the same way it may be shewn that AD is parallel to BC, and .. ABCD is a parallelogram. (3.) To bisect a parallelogram by a line drawn from a point in one of its sides. Let ABCD be a parallelogram, and P a given point in the side AB. Draw the diameter BD, which bisects the parallelogram. Bisect BD E F in F; join PF, and produce it to E. PE bisects the parallelogram. Since the angle PBD is equal to the angle BDE, and the vertically opposite angles at Fare equal, and BF =FD, .. the triangles PBF, DFE are equal. But the triangle ABD is equal to BDC, .. APFD is equal to BFEC; and to these equals adding the equal triangles DFE, PFB, the figure APED - PECB; and AC is.. bisected by PE. = COR. Any line drawn through the middle point of the diameter of a parallelogram is bisected in that point. (4.) If from any point in the diameter (or diameter produced) of a parallelogram straight lines be drawn to the opposite angle; they will cut off equal triangles. Q From any point E, in AC the diameter of the parallelogram ABCD, let lines EB, A E D ED be drawn; the triangles ABE, AED are equal; as also the triangles BEC, CED. Draw the diameter BD. The bases BF, FD being equal, the triangles BFA, DFA (Eucl. i. 38.), as also the triangles BFE, DFE are equal, hence .. BAE, DAE are equal. And ABC being equal to ADC, the triangles BEC, DEC are also equal. (5.) From one of the angles of a parallelogram to draw a line to the opposite side, which shall be equal to that side together with the segment of it which is intercepted between the line and the opposite angle. Let ABCD be the parallelogram, A the angle from which the line is to be drawn. Produce DC to E, making CE=CD. Join AE, and at the point A make the angle EAF-AEF; AF is the line required. A B C F E For CE being equal to CD, EF is equal to DC and CF together; and the angles FEA, FAE being equal, FA= FE, and .. AF= DC and CF together. COR. In the same manner if CE = CB, AF=EF= BC and CF together. (6.) If from one of the angles of a parallelogram a straight line be drawn cutting the diameter, a side, and a side produced; the segment intercepted between the angle and the diameter, is a mean proportional between the segments intercepted between the diameter and the sides. From B, one of the angles of the parallelogram ABCD, let any line BE be drawn cutting the diameter AC in F, the opposite side in G, and A B T Η D E AD produced in E; BF is a mean proportional between FG and FE. Draw DH parallel to BF, and .. equal to it; ... also AH-FC, and AF-CH. Since HD is parallel to BG, FG DH (:: CF: CH: AH: AF) :: DH: FE, or FG BF :: BF: FE. (7.) The two triangles, formed by drawing straight lines from any point within a parallelogram to the extremities of two opposite sides, are together half of the parallelogram. Let P be any point within the parallelogram ABCD, from which let lines PA, PD, PB, PC be drawn to the extremities of the opposite sides; the triangles PAD, D F E PBC are equal to half the parallelogram; as also the triangles APB, DPC. Through E draw EPF parallel to AD or BC; then (Eucl. i. 41.) the triangle APD is half of AEFD, and BPC is half of BEFC, .. APD, BPC are together half |