7.5: Parallelograms
 Page ID
 23623
A quadrangle \(ABCD\) in the Euclidean plane is called nondegenerate if no three points from \(A, B, C, D\) lie on one line.
A nondegenerate quadrangle is called a parallelogram if its opposite sides are parallel.
Lemma \(\PageIndex{1}\)
Any parallelogram is centrally symmetric with respect to a midpoint of one of its diagolals.
In particular, if \(\square ABCD\) is a parallelogram, then
(a) its diagonals \([AC]\) and \([BD]\) intersect each other at their midpoints;
(b) \(\measuredangle ABC = \measuredangle CDA\);
(c) \(AB = CD\).
 Proof

Let \(\square ABCD\) be a parallelogram. Denote by \(M\) the midpoint of \([AC]\).
Since \((AB)\parallel (CD)\), Theorem 7.2.1 implies that \((CD)\) is a reflection of \((AB)\) across \(M\). The same way \((BC)\) is a reflection of \((DA)\) across \(M\). Since \(\square ABCD\) is nondegenerate, it follows that \(D\) is a reflection of \(B\) across \(M\); in other words, \(M\) is the midpoint of \([BD]\).
The remaining statements follow since reflection across \(M\) is a direct motion of the plane (see Proposition 7.2.1).
Exercise \(\PageIndex{1}\)
Assume \(ABCD\) is a quadrangle such that
\(AB = CD = BC = DA.\)
Such that \(ABCD\) is a parallelogram.
 Hint

Since \(\triangle ABC\) is isosceles, \(\measuredangle CAB = \measuredangle BCA\).
By SSS, \(\triangle ABC \cong \triangle CDA\). Therefore, \(\pm \measuredangle DCA = \measuredangle BCA = \measuredangle CAB\).
Since \(D \ne C\), we get "" in the last formula. Use the transversal property (Theorem 7.3.1) to show that \((AB) \parallel (CD)\). Repeat the argument to show that \((AD) \parallel (BC)\).
A quadrangle as in the exercise above is called a rhombus.
A quadrangle ABCD is called a rectangle if the angles ABC, BCD, CDA, and DAB are right. Note that according to the transversal property (Theorem 7.3.1), any rectangle is a parallelogram.
A rectangle with equal sides is called a square.
Exercise \(\PageIndex{2}\)
Show that the parallelogram \(ABCD\) is a rectangle if and only if \(AC = BD\).
 Hint

By Lemma \(\PageIndex{1}\) and SSS, \(AC = BD\) if and only if \(\angle ABC = \pm \measuredangle BCD\). By the transversal property (Theorem 7.3.1), \(\measuredangle ABC + \measuredangle BCD \equiv \pi\).
Therefore, \(AC = BD\) if and only if \(\measuredangle ABC = \measuredangle BCD = \pm \dfrac{\pi}{2}\).
Exercise \(\PageIndex{3}\)
Show that the parallelogram \(ABCD\) is a rhombus if and only if \((AC) \perp (BD)\).
 Hint

Fix a parallelogram \(ABCD\). By Lemma \(\PageIndex{1}\), its diagonals \([AC]\) and \([BD]\) have a common midpoint; denote it by \(M\).
Use SSS and Lemma \(\PageIndex{1}\) to show that
\(AB = CD \Leftrightarrow \triangle AMB \cong \triangle AMD \Leftrightarrow \measuredangle AMB = \pm \dfrac{\pi}{2}.\)
Assume \(\ell \parallel m\), and \(X, Y \in m\). Let \(X'\) and \(Y'\) denote the foot points of \(X\) and \(Y\) on \(\ell\). Note that \(\square XYY'X'\) is a rectangle. By Lemma \(\PageIndex{1}\), \(XX' = YY'\). That is, any point on \(m\) lies on the same distance from \(\ell\). This distance is called the distance between \(\ell\) and \(m\).