7.1: Parallel lines
( \newcommand{\kernel}{\mathrm{null}\,}\)
In consequence of Axiom II, any two distinct lines
To emphasize that two lines on a diagram are parallel we will mark them with arrows of the same type.
Let
- Proof
-
Assume the contrary; that is,
. Then there is a point, say , of intersection of and . Then by Theorem 5.3.1, . Since any line is parallel to itself, we have that — a contradiction.
For any point
The above theorem has two parts, existence and uniqueness. In the proof of uniqueness we will use the method of similar triangles.
- Proof
-
Apply Theorem 5.3.1 two times, first to construct the line
thru that is perpendicular to , and second to construct the line thru that is perpendicular to . Then apply Proposition .Uniqueness. If
, then by the definition of parallel lines. Further we assume .Let us construct the lines
and as in the proof of existence, so .Assume there is yet another line
parallel to . Choose a point taht lies with on the same side from . Let be the foot point of on .Let
be the point of intersection of and . According to Proposition . Therefore, , and lie on the same side of . In particular, .Choose
such thatBy SAS similarity condition (or equivalently by Axiom V) we have that
; therefore . It follows that lies on and - a condition.
Assume
- Proof
-
Assume the contrary; that is,
. Then there is a point . By Theorem , -- a contradiction.
Note that from the definition, we have that
(i)
(ii) if
(iii) if
Let
- Hint
-
Apply Proposition
to show that . By Corollary , . The latter contradicts that .
Make a ruler-and-compass construction of a line thru a given point that is parallel to a given line.
- Hint
-
Repeat the construction in Exercise 5.7.1 twice.