VI. ON PARALLELOGRAMS. — ON TRANSLATIONS 47
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
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.
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
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