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 A 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 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 angles; they will cut off equal triangles. Q From any point E, in AC the diameter of the parallelo gram ABCD, let lines EB, Α 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. = E For CE being equal to CD, EF = 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 and 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. AD produced in E; BF is a mean proportional between Draw DH parallel to BF, and ... equal to it ; .. also (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 A F PBC are equal to half the parallelogram; as also the Through E draw EPF parallel to AD or BC; then of ABCD. In the same manner if a line be drawn through P parallel to AB or DC, it may be shewn that APB, DPC together are half of ABCD. (8.) If a straight line be drawn parallel to one of the sides of a parallelogram, and one extremity of this line be joined to the opposite one of the parallel side, by a line which also cuts the diameter; the segments of the diameter made by this line will be reciprocally proportional to the segments of that part of it which is intercepted between the side and the parallel line. Let EF be drawn parallel to AD one of the sides of the parallelogram ABCD, cutting the diameter BD in G. Join AF, cutting it also in H; then will BH : HD :: HD: HG. For the angle ABH being equal to HDF, and AHB DHF, the triangles AHB, DHF are equiangular, and .. BH: HD :: AH : HF :: DH: HG, since the triangles AHD, FHG are also equiangular. (9.) If two lines be drawn parallel and equal to the adjacent sides of a parallelogram; the lines joining their extremities, if produced, will meet the diameter in the same point. Let HI, FG be drawn equal and parallel to the adjacent sides AB, BC of the parallelogram ABCD. Join HF, GI; these lines produced will meet the diameter DB in the same point. |