3 The parallel axiom

Axiom 8 (Parallel Axiom)   Given a line $k$, and a point $A$ not on $k$, there is exactly one line $m$ passing through $A$ and parallel to $k$.

We remark that the point of the axiom is not the existence of the parallel, but the uniqueness. We will see below that existence actually follows from what we already know.

It is sometimes convenient to think of a line as being parallel to itself, so we make the following formal definition. Two lines are not parallel if they have exactly one point in common; otherwise they are parallel.

Theorem 3.1   In the set of all lines in the plane, the relation of being parallel is an equivalence relation.

Proof. First, since a line has infinitely many points in common with itself, it is parallel to itself; hence the relation is reflexive (this is the point of the strange definition).

Second, the definition is obviously symmetric; it is defined in terms of the two lines; not one with relation to the other.

Third, suppose $k$ is parallel to $m$, and $m$ is parallel to $n$. There is obviously nothing further to prove unless the three lines are distinct. Assume that $k$ and $n$ are not parallel. Since two lines are either equal, parallel, or have exactly one point in common, we must have that $k$ and $n$ have a point in common. But this contradicts Axiom 8. $\qedsymbol$

3.1 Alternate interior angles

We will meet the following situation some number of times. We are given two lines $k_1$ and $k_2$, and a third line $m$, where $m$ crosses $k_1$ at $A_1$ and $m$ crosses $k_2$ at $A_2$. Choose a point $B_1\neq A_1$ on $k_1$, and choose a point $B_2\neq A_2$ on $k_2$, where $B_1$ and $B_2$ lie on opposite sides of the line $m$. Then $\angle B_1A_1A_2$ and $\angle B_2A_2A_1$ are referred to as alternate interior angles.

In any given situation, there are two distinct pairs of alternate interior angles. That is, let $C_1$ be some point on $k_1$, where $B_1$ and $C_1$ lie on opposite sides of $m$, and let $C_2$ be some point on $k_2$, where $C_2$ and $B_2$ lie on opposite sides of $m$. Then one could also regard $\angle C_1A_1A_2$ and $\angle C_2A_2A_1$ as being alternate interior angles. However, observe that $m\angle B_1A_1A_2+m\angle C_1A_1A_2=\pi$ and $m\angle B_2A_2A_1+m\angle C_2A_2A_1=\pi$. It follows that one pair of alternate interior angles are equal if and only if the other pair of alternate interior angles are equal.

Proposition 3.2   If the alternate interior angles are equal, then the lines $k_1$ and $k_2$ are parallel.

Proof. Suppose not. Then the lines $k_1$ and $k_2$ meet at some point $D$. If necessary, we interchange the roles of the $B_i$ and the $C_i$ so that $\angle B_1A_1A_2$ is an exterior angle of $\triangle A_1A_2D$. Then $D$ and $B_2$ lie on the same side of $m$, so $\angle DA_2A_1=\angle B_2A_2A_1$. By the exterior angle inequality,

$$m\angle B_1A_1A_2>m\angle A_1A_2D=m\angle B_2A_2A_1=m\angle B_1A_1A_2,$$

so we have reached a contradiction. $\qedsymbol$

3.2 Existence of parallel lines

Let $k_1$ be a line, and let $A_2$ be a point not on $k_1$. Pick some point $A_1$ on $k_1$ and draw the line $m$ through $A_1$ and $A_2$. By Axiom 7, we can find a line $k_2$ through $A_2$ so that the alternate interior angles are equal. Hence we can find a line through $A_2$ parallel to $k_1$.

Theorem 3.3 (alternate interior angles equal)   Two lines $k_1$ and $k_2$ are parallel if and only if the alternate interior angles are equal.

Proof. To prove the forward direction, construct the line $k_3$ through $A_2$, where there is a point $B_3$ on $k_3$, with $B_3$ and $B_2$ on the same side of $m$, so that $m\angle B_3A_2A_1=m\angle B_1A_1A_2$. Then, by Prop. 3.2, $k_3$ is a line through $A_2$ parallel to $k_1$. Axiom 8 implies $k_3=k_1$. Hence $m\angle B_3A_2A_1=m\angle B_2A_2A_1$, and the desired conclusion follows.

The other direction is just Prop. 3.2, restated as part of this theorem for convenience. $\qedsymbol$

3.3 The sum of the angles of a triangle

Theorem 3.4   The sum of the measures of the angles of a triangle is equal to $\pi$.

Proof. Consider $\triangle ABC$, and let $m$ be the line passing through $A$ and parallel to $BC$. Let $D$ and $E$ be two points on $m$, on opposite sides of $A$, where $D$ and $C$ lie on opposite sides of the line $AB$. Then $B$ and $E$ lie on opposite sides of $AC$.

Exercise 3.1   Use alternate interior angles to complete the proof of this theorem.

$\qedsymbol$

A quadrilateral is a region bounded by four line segments, so it has four verticies on its boundary.

Corollary 3.5   The sum of the measures of the angles of a quadrilateral is $2\pi$.

Proof. Cut the triangle into two triangles, and do the obvious computation. $\qedsymbol$

A rectangle is a quadrilateral in which all four angles are right angles.

Theorem 3.6   If $ABCD$ is a rectangle, then $AB$ is parallel to $CD$, and $\vert AB\vert=\vert CD\vert$. Similarly, $BC$ is parallel to $AD$ and $\vert BC\vert=\vert AD\vert$.

Exercise 3.2   Prove this theorem.

i.

Prove that opposite pairs of sides are parallel.

ii.

Now cut the rectangle into two triangles; prove that these two triangles are congruent. Conclude that opposite sides of the rectangle have equal length.

Somewhat more generally, a parallelogram is a quadrilateral $ABCD$ in which opposite sides are parallel; that is, $AB$ is parallel to $CD$, and $AD$ is parallel to $BC$.

A rectangle with all four sides of equal length is a square; a parallelogram with all four sides of equal length is a rhombus.

Theorem 3.7   Let $ABCD$ be a parallelogram. Then $m\angle A=m\angle C$; $m\angle B=m\angle D$; $\vert AB\vert=\vert CD\vert$; and $\vert BC\vert=\vert AD\vert$.

Exercise 3.3   Prove this theorem. (Hint: Draw a diagonal.)

Theorem 3.8   If $ABCD$ is a quadrilateral in which $\vert AB\vert=\vert CD\vert$ and $\vert AD\vert=\vert BC\vert$, then $ABCD$ is a parallelogram.

Exercise 3.4   Prove this theorem.

Theorem 3.9   Let $ABCD$ be a parallelogram with diagonals of equal length (that is, $\vert AC\vert=\vert BD\vert$). Then $ABCD$ is a rectangle.

Exercise 3.5   Prove this theorem.