Results related to progress report 5.2

In this section we make the connection between poles and the elliptic parallel postulate.

Theorem

If $P$ is a pole of $l$ and $m$ is perpendicular to $l$, then $P$ is on $m$.

Proof

Suppose that $u$ and $v$ are both perpendicular to $l$ and that they meet at $P$. Since $u$ and $v$ are fixed by $R_{l}$, we have that $P$ is a fixed point of $R_{l}$. Now suppose that $m$ meets $l$ at a point $Q$. Since $P $ and $Q$ are both fixed points of $R_{l}$, it follows from the results of the preceding section that MATH is a fixed line of $R_{l}$ and thus, by the fixed line theorem, MATH is perpendicular to $l$ (as $P$ is not on $l$). Since MATH and $m$ are both perpendicular to $l$ and both meet $l$ at $Q$, the uniqueness of perpendiculars yields that MATH and thus $P$ is on $m$.

Theorem

If $P$ is a pole of $l$ and $P$ is on a line $m$, then $l$ is perpendicular to $m$.

Proof

Select a point $Q$ on $m$ and drop a perpendicular $m^{\prime }$ from $Q$ onto $l$. The previous result shows that $P$ is on $m^{\prime }$. As $m$ and $m^{\prime }$ both contain $P$ and $Q$, $m=m^{\prime }$ and hence $m$ is perpendicular to $l$.

Theorem

A line has at most one pole.

Proof

Suppose $P$ and $\ Q$ are poles of a line $l$. Suppose that $u$ and $v$ are two lines perpendicular to $l$ that meet at $P$. The first result of this section shows that $Q$ is on both $u$ and $\ v$. As $u$ and $v$ can meet in at most one point, $P=Q$.

Theorem

If one line has a pole, then every line has a pole.

Proof

Suppose that $l$ is a line with a pole $P$ and that $m$ is a line. We wish to show that $m$ has a pole. There are two cases to consider: $m$ is perpendicular to $l$ and $m$ is not perpendicular to $l$.

First we consider the case $m$ is perpendicular to $l$. Note that $P$ is on $m$ and construct a line $c$ perpendicular to $m$ passing through $P$. Since $P$ is on $c$, it follows that $c$ is perpendicular to $l$ and hence $c $ and $l$ must intersect; let $S$ denote this point of intersection. The point $S$ is the pole of $m$.

Note that the above argument shows that any line perpendicular to a line with a pole also has a pole.

Next suppose that $m$ is not perpendicular to $l$. In this case $P$ is not on $m$. Drop a perpendicular $d$ from $P$ to $m$; observe that since $P$ is on $d$, the line $d$ is perpendicular to $l$. Now construct a line $e$ that contains $P$ and is perpendicular to $d$. Let $T$ denote the point of intersection of $e$ and $l$ and observe that $T$ is the pole of the line $d$. Now, since $m$ is perpendicular to $d$ and $d$ has a pole, $m$ has a pole.

Theorem

Given lines $l$ and $m$ with poles $P$ and $Q$. The $l$ and $m$ intersect at the pole of the line MATH.

Proof

First note that since one line has a pole, then every line has a pole and hence MATH has a pole. Now note that $l$ and $m$ are both perpendicular to MATH (as $P$ is on MATH and $Q$ is on MATH). Thus $l$ and $m$ meet at the pole of MATH.





Lecture: Theorems on Three Reflections

Theorem

(First theorem on three reflections). If the lines $l,m,$ and $n$ intersect at a point $P$, then there is a line $w$ passing through $P$ such that MATH.

Proof

First note that if $m=n$, then MATH and the conclusion of the theorem follows. Thus we may assume, without loss of generality, that $m\neq n$. Now note that axiom M.3 guarantees that there is a line $d$ such that MATH for every point $P$ on $l$ and that $m,n$ and $d$ are concurrent Note that this implies that MATH for every point $P$ on $l.$ As $R_{d}R_{n}R_{m}$ is an orthogonal collineation, it follows that either MATH or $R_{d}R_{n}R_{m}=I$.

Observe that if MATH, then it follows immediately that MATH and the conclusion of the theorem is satisfied.

To complete the proof, then, it suffices to show that $R_{d}R_{n}R_{m}=I$ leads to a contradiction. If $R_{d}R_{n}R_{m}=I$, then MATH and MATH MATH and hence MATH. Thus $d$ is perpendicular to $n$ (note that if $n=d$, then $R_{m}=I,$ a contradiction). A similar calculation can be used to show that $m$ is perpendicular $d$. As there is only one perpendicular to $d$ at the point where $m,n,$ and $d$ intersect, it follows that $m=n$. This, however, contradicts the assumption that $m\neq n$ and hence MATH.

Theorem

If $M=R_{l}R_{m}$ is a rotation about $P$, then given any line $u$ passing through $P$ there is another line $v$ passing through $P$ such that $M=R_{v}R_{u}$.

Proof

Let $l,m,$ and $u$ be as in the statement of the theorem. The first theorem on three reflections yields that there is a line $v$ passing through $P$ such that MATH. It is immediate that MATH.

Theorem

(Second theorem on three reflections). If the lines $l,m,$ and $n$ are all perpendicular to a line $t$, then there is a line $z$ perpendicular to $t$ such that MATH.

Proof

This proof is very similar to Theorem 1.

First note that if $m=n$, then MATH and the conclusion of the theorem follows. Thus we may assume, without loss of generality, that $m\neq n$. Now note that axiom M.4 guarantees that there is a line $d$ such that MATH.MATH for every point $P $ on $l$ and that $m,n$ and $d$ are perpendicular $t$. Note that this implies that MATH for every point $P$ on $l.$ As $R_{d}R_{n}R_{m}$ is an orthogonal collineation, it follows that either MATH or $R_{d}R_{n}R_{m}=I$.

Observe that if MATH, then it follow immediately that MATH and the conclusion of the theorem is satisfied.

To complete the proof, then, it suffices to show that $R_{d}R_{n}R_{m}=I$ leads to a contradiction. If $R_{d}R_{n}R_{m}=I$, then MATH and MATH MATH and hence MATH. Thus $d$ is perpendicular to $n$ (note that if $n=d$, then $R_{m}=I,$ a contradiction). A similar calculation can be used to show that $m$ is perpendicular $d$. Note that $m,n,$and $d$ intersect at the pole of $t.$ As there is only one perpendicular to $d$ at the point where $m,n,$ and $d$ intersect, it follows that $m=n$. This, however, contradicts the assumption that $m\neq n$ and hence MATH.

Note that it is not enough to show that $m$ is perpendicular to $d$ to arrive at a contradiction; it could well be that $m$ and $d$ meet at the pole of $t$.

Theorem

If $T=R_{l}R_{m}$ is a translation along $t$ and $u$ is a line perpendicular to $t$, then there is a line $v$ perpendicular to $t$ such that $T=R_{v}R_{u} $.

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