46 VI. ON PARALLELOGRAMS. — ON TRANSLATIONS
In this theorem, the hypothesis consists of two parts:
1◦. two sides are parallel;
2◦. the other two sides are also parallel.
The conclusion also has two parts:
1◦. two opposite sides are equal;
2◦. the other two sides are also equal.
Since we can form a converse by using either the whole or a part of the original
conclusion, and vice versa, this theorem has two converses.
Converses. A quadrilateral is a parallelogram:
If the opposite sides are equal;
If two opposite sides are parallel and equal.
Suppose that AB = CD and AD = BC in quadrilateral ABCD (Fig. 44).
We again draw diagonal AC. Triangles ABC, CDA will be congruent, since they
have three sides equal in pairs. Thus angles A1 and C1 are equal, and since these
are alternate interior angles with respect to transversal AC, lines AB, CD must be
parallel. Likewise, the equality of A2, C2 proves that AD and BC are parallel.
Assume now that AB = CD and AB is parallel to CD. Triangles ABC,
CDA are again congruent, because they have an equal angle (A1 = C1 are alternate
interior angles) between two pairs of equal sides: AC is a common side and AB =
CD. From this congruence we again ﬁnd that A2 = C2, and that sides AD, BC
46b. Remark I. A quadrilateral can have two sides AB, CD equal, and the
other two BC, AD parallel, without being a parallelogram (it is then called an
Choosing an arbitrary side AB, it suﬃces to take the line symmetric to AB
(D1C, Fig. 45) with respect to any line xy in the plane, provided that the line xy is
not parallel to AB, and that xy intersects the line AB in a point I on the extension
of AB (and not on segment AB itself). Then quadrilateral ABCD1 will have two
parallel sides (both perpendicular to xy) and the other two equal (because they
are symmetric to each other). These last two sides are not parallel because they
intersect at point I.
Conversely, every quadrilateral with two sides BC, AD parallel, and the other
two equal is either a parallelogram, or (in the case of an isosceles trapezoid) a
ﬁgure with a line of symmetry. Indeed, let xy be the perpendicular bisector of BC.
We have already found two oblique lines from C to the line AD, both equal to
AB, namely: CD1 which is symmetric to AB with respect to xy, and CD2 which,