VI. ON PARALLELOGRAMS. — ON TRANSLATIONS 47 Figure 45 together with A, B, and C forms a parallelogram. These oblique lines certainly have endpoints on AD, since that line is parallel to BC. It follows that D must be either D1 or D2, since there are only two oblique lines equal to AB from C to AD. The reasoning above may seem to be incorrect if D2 coincides with D1 that is, when CD1 is parallel to AB. This requires, as we have just seen, that AB also be parallel to xy, and therefore perpendicular to AD, and indeed in this case there is only one oblique line from C with length equal to AB (the perpendicular). Remark II. In the first part of the preceding converse it is essential that the quadrilateral be proper (21, Remark): it is only if the triangles ABC, ADC of Fig. 44 are on different sides of the common side AC that the angles A1, C1 are alternate interior. It is easy to construct an improper quadrilateral (called a anti-parallelogram) whose opposite sides are equal. It suﬃces, in parallelogram ABCD (Fig. 45b), to replace the point D by point E symmetric to it with respect to the diagonal AC. Figure 45b We can also obtain an anti-parallelogram ABCE1 by taking E1 to be symmetric to B with respect to the perpendicular bisector of AC, so that ABE1C is an isosceles trapezoid. But this point E1 is none other than E, because (24, case 3◦) there is, on any one side of AC, only one point E which is both at a distance AE = BC from A and also at a distance CE = AB from C. Every improper quadrilateral

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2008 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.