Proving Lines Parallel or Perpendicular

First run the argument with the two $57^\circ$ corresponding angles, $\angle2$ and $\angle6$. Through $P$ there is exactly one parallel $n$ to $m$, by the accepted parallel foundation. Its corresponding angle with $t$ is $57^\circ$, by the theorem from the previous lesson. The given line $\ell$ makes that same $57^\circ$ opening on the same side of the same ray of $t$. The two rays must coincide, so $\ell$ is $n$ and is parallel to $m$.

Let distinct coplanar lines $\ell$ and $m$ be cut by $t$ at distinct points $P$ and $Q$. Suppose their corresponding angles $\angle2$ and $\angle6$ are congruent. Their common measure lies strictly between $0^\circ$ and $180^\circ$, since these are intersections of distinct lines.

Assume for contradiction that $\ell$ is not parallel to $m$. Point $P$ is outside $m$, because $P$ and $Q$ are distinct and $t$ meets $m$ only at $Q$. The accepted existence and uniqueness of a parallel gives a line $n$ through $P$ parallel to $m$.

Since $n\parallel m$ while the temporary assumption says $\ell$ is not parallel to $m$, the lines $n$ and $\ell$ are different. Through $P$, they therefore give different rays on the chosen side of $t$. The forward corresponding-angle theorem gives the selected angle between $n$ and the upward ray of $t$ the same measure as $\angle6$. The given congruence gives $\angle2$ that measure too. These would be two different rays on the same chosen side making the same opening with the same starting ray, contradicting the accepted angle-measure foundation. Reject the temporary assumption, so $\ell\parallel m$.

The value $57^\circ$ did no work beyond making the shared opening concrete. This argument establishes the converse for any valid corresponding-angle measure and any corresponding position.

For example, suppose $\angle4$ and $\angle6$ both measure $68^\circ$. Vertical angles make $m\angle2=68^\circ$, so corresponding angles $2$ and $6$ are congruent and $\ell\parallel m$. Or suppose same-side interior angles $4$ and $5$ measure $71^\circ$ and $109^\circ$. The lower linear pair gives $m\angle6=180^\circ-109^\circ=71^\circ$, reducing to the preceding test.

Keep distinct coplanar lines cut by one transversal at distinct points. If alternate interior angles $4$ and $6$ are congruent, vertical angles give $\angle2\cong\angle4$. Transitivity gives $\angle2\cong\angle6$, so the corresponding-angle converse gives parallelism. The other alternate interior pair follows with the labels exchanged.

If same-side interior angles $4$ and $5$ are supplementary, their measures add to $180^\circ$. The linear pair $5,6$ also sums to $180^\circ$. Subtracting $m\angle5$ from these equal sums gives $m\angle4=m\angle6$. The alternate interior converse just proved gives parallelism. The other side follows the same way.

If alternate exterior angles $1$ and $7$ are congruent, vertical angles at $Q$ give $\angle7\cong\angle5$. Then $\angle1\cong\angle5$, a corresponding pair, so the corresponding converse gives parallelism. The other alternate exterior pair works identically.

If same-side exterior angles $1$ and $8$ are supplementary, the linear pair $5,8$ also sums to $180^\circ$. Subtracting $m\angle8$ gives $m\angle1=m\angle5$, again a corresponding pair. The other same-side exterior pair has the same argument.

Let $\ell\parallel m$ be distinct lines in one plane, and let a third coplanar line $n$ cross $m$ at $R$. Suppose $n$ did not meet $\ell$. These distinct coplanar lines would then be parallel. Since $R$ lies on $m$, it is outside $\ell$. Both $m$ and $n$ would be different parallels to $\ell$ through $R$, contradicting the accepted uniqueness of a parallel through an external point. Therefore $n$ meets $\ell$. The two crossings are distinct because $\ell$ and $m$ do not meet, so $n$ is a transversal.

Let $\ell$ and $m$ be distinct coplanar lines perpendicular to $n$. They cannot meet $n$ at the same point: on either chosen side of $n$, the unique ray making a $90^\circ$ opening would force $\ell$ and $m$ to coincide. Their feet on $n$ are therefore distinct. Line $n$ is a transversal, and the corresponding right angles are congruent, so the corresponding-angle converse gives $\ell\parallel m$. The coplanar condition is essential; spatial lines can have different directions while each meets a third line at a right angle.

Now let three pairwise distinct coplanar lines $\ell,m,n$ satisfy $\ell\parallel n$ and $m\parallel n$. Suppose $\ell$ and $m$ met at $R$. Point $R$ would lie outside $n$, since neither parallel meets $n$. There would then be two different parallels to $n$ through $R$, contradicting the accepted uniqueness of a parallel. Thus $\ell$ and $m$ do not meet. As distinct coplanar lines, they are parallel.

Let $P$ be outside line $\ell$ in the same plane. The accepted parallel foundation gives a line $p$ through $P$ parallel to $\ell$. At $P$, the accepted angle-measure foundation with measure $90^\circ$ gives a ray perpendicular to $p$; extend it to a full line $q$. The crossing result shows that $q$ meets $\ell$, and the perpendicular transversal theorem makes that meeting perpendicular. Thus a foot exists.

If two different perpendicular lines to $\ell$ passed through $P$, the proved two-perpendiculars result would make them parallel. They would also meet at $P$, a contradiction. Hence the perpendicular line and its foot are unique. This proves existence and uniqueness; the compass-and-straightedge construction comes in lesson 9.1.