
# 7.5: Parallelograms


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 $$m$$.