Download Feb 23 Notes: Definition: Two lines l and m are parallel if they lie in

Feb 23 Notes:
Definition: Two lines l and m are parallel if they lie in the same
plane and do not intersect.
Terminology: When one line intersects each of two given lines,
we call that line a transversal.
We define alternate interior angles, corresponding angles,
alternate exterior angles, and interior angles on the same
side of the transversal using various betweeness and half-plane
notions.
Suppose line l intersects lines m and n at points B and E,
respectively, with points A and C on line m and points D and F
on line n such that A-B-C and D-E-F, with A and D on the same
side of l. Suppose also that G and H are points such that H-E-BG.
Then pABE and pBEF are alternate interior angles, as are
pCBE and pDEB.
pABG and pFEH are alternate exterior angles, as are pCBG
and pDEH.
pGBC and pBEF are a pair of corresponding angles, as are
pGBA & pBED, pCBE & pFEH, and pABE & pDEH.
pCBE and pFEB are interior angles on the same side of the
transversal, as are pABE and pDEB.
Our Last Theorem in Absolute Geometry: If two lines in the
same plane are cut by a transversal so that a pair of alternate
interior angles are congruent, the lines are parallel.
Proof: Let l intersect lines m and n at points A and B
respectively. Let p1– p2. Suppose m and n meet at point C.
Then either p1 is exterior to ªABC, or p2 is exterior to ªABC.
In the first case, the exterior angle inequality gives p1 > p2; in
the second, it gives p2 > p1. In either case, we have a
contradiction to p1– p2.
The converse of this statement is one way of stating the
Euclidean Parallel Postulate. There are many other
formulations as well. Here is a list of a few statements
equivalent to Parallel Postulate.
1.
If two parallel lines are cut by a transversal, alternate interior
angles are congruent.
2.
If two lines in the same plane are cut by a transversal so that
the sum of the measures of a pair of interior angles on the
same side of the transversal is less than 180, then the lines
will meet on that side of the transversal.
3.
A third line intersecting one of two parallel lines intersects
the other.
4.
A line perpendicular to one of two parallel lines is
perpendicular to the other.
5.
The perpendicular bisectors of the sides of a triangle are
concurrent.
6.
There exists a circle passing through any three noncollinear
points.
7.
The sum of measures of the angles of a triangle is 180.
8.
There exists one triangle such that the sum of measures of the
angles is 180.
9.
There exists a rectangle.
10. There exist two lines l and m such that l is equidistant from m.
11. There exists a pair of similar, noncongruent triangles.
12. There exists an acute angle such that every line intersecting
and perpendicular to one ray of the angle intersects the other
ray.
We take as our version of the Parallel Postulate the following:
Axiom P-1: If l is any line and P any point not on l, there exists in
the plane of l and P one and only one line m that passes through P
and is parallel to l.
We can now prove Statement 1 on our list,
namely:
Theorem: If two parallel lines are cut by a transversal, alternate
interior angles are congruent.
Proof: Let l and m be parallel lines cut by a transversal t at points
Q and P, respectively. Suppose for contradiction that alternate
interior angles p1 (at point P) and p2 (at point Q) are not
congruent, and WLOG that mp1 > mp2. Using the angle
construction theorem, create ray
on the other side of t from p2
such that mpRPQ = mp2. Then from the above theorem, since
pRPQ and p2 are alternate interior angles, it follows that lines l
and
are parallel. But that means there are two lines through P
parallel to line l, contradicting the parallel postulate. So p1 and
p2 are congruent.
Now comes a flood of straightforward corollaries. First, notice
that, given two lines in the same plane cut by a transversal,
The following are equivalent:
1.
2.
3.
4.
5.
A pair of alternate interior angles are congruent
Both pairs of alternate interior angles are congruent.
Both pairs of alternate exterior angles are congruent
All pairs of corresponding angles are congruent.
Both pairs of interior angles lying on one side of the
transversal are supplementary.
This is very easy to prove using vertical angles and linear pairs so
we’ll leave the proof as an exercise. From these equivalences
come the C, F, and Z properties listed in your text on page 215.
Taken together with the above two theorems, we have the
following:
If two lines in the same plane are cut by a transversal, then the
lines are parallel iff
•
•
•
•
•
alternate interior angles are congruent
alternate exterior angles are congruent
corresponding angles are supplementary
interior angles on the same side of the transversal are
supplementary
exterior angles on the same side of the transversal are
supplementary
An immediate consequence is:
If a line is perpendicular to one of two parallel lines, it is
perpendicular to the other also.
Euclidean Exterior Angle Theorem: In any triangle, the measure of
an exterior angle is the sum of the measures of the two remote
interior angles.
Proof: We use the same construction as for the proofs of the
exterior angle theorem and Saccheri - Legendre Theorem in
absolute geometry. Begin with ªABC and point D with B-C-D.
Find the midpoint M of
and find point E with B-M-E and MB
= ME. Connect E and C. Now by construction and SAS, ªBAM
is congruent to ªECM, and it is easy to show that E is interior to
pACD. Moreover, since pBAC –pECA, with
as the
transversal we have congruent alternate interior angles and thus
. This makes pABC and pECD corresponding angles
(with
as the transversal), and hence congruent. Thus
Another flood of corollaries:
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•
•
•
•
The sum of the measure of the angles of any triangle is 180.
The acute angles of a right triangle are complementary
The sum of the measures of any convex quadrilateral is 360
Rectangles exist; in fact, every Saccheri quadrilateral and
every Lambert quadrilateral is a rectangle
Squares exist.
This sharpens most of the “almost Euclidean” results we’ve gotten
in the last few chapters. We have one left:
Midpoint Connector Theorem: The segment joining the
midpoints of two sides of a triangle is parallel to the third side and
has length one-half the third side.
Proof: Begin with ª ABC with L and M the midpoints of
and
respectively. Extend
to find P with L-M-P and LM =
MP. Connect P and C. By construction and SAS, ªLAM – ªPCM
and so
(pLAC –pPCA, alternate interior angles). Note
also that BL = LA = PC. If we construct
, we have pBLC
–pPCL (alternate interior angles with
as the transversal) so
we get ªBLC – ªPCL by SAS. Thus
by
CPCF, and since pCLP – pLCB, and they are alternate interior
angles with
as the transversal,
.
Corollary: If a line bisects one side of a triangle and is parallel to
the second, it also bisects the third side. (Follows from uniqueness
of parallels.)