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Chapter 2
EXISTENCE AND UNIQUENESS
One of the basic axioms of Euclidean
geometry says that two points
determine a unique line.
This implies that two distinct lines cannot
intersect in two or more points, they
can either intersect in only one point or
not at all.
Two lines that don’t intersect are
called parallel.
PROBLEM
l
Given a line l and a point P not on ‚
construct a line through P and parallel to l .
L
let A be any point on l, and draw AP
Then draw a line PQ so that
QPA
 PAB
as
shown
in the figure. This will be
the desired line.
The proof will be by contradiction.
If PQand l are not parallel, we may assume
without loss of generality that they intersect
as in the figure on the side of B at the point C.
Now consider PAC. the exterior angle APQ is
equal to the interior angle PAC. But this
contradicts the exterior angle theorem, which
states that QPA  PAC .
Hence PQ must be parallel to l .
I
Given a line l and a point P not on l ,
there exists a line that contains P
and is parallel to l .
COROLLARY
Given lines ABand PQ as in the figure,
if QPA  PAB, then
to PQ .
AB is parallel
THE PARALLEL POSTULATE
If
l is any line and P is a point not on l .
then there is no more than one line through P
parallel to l .
Opposite Interior Angles Theorem
Let AB and PQ be parallel lines with
transversal AP such that QPA and PAB
are opposite interior angles.
Then
QPA  PAB
.
The proof will be by contradiction. If the theorem
was false and if QPA  PAB , then we could
construct a distinct line PQ through P such that
APQ  PAB. Since APQ and PAB
are opposite interior angles, their congruence
implies that PQ
AB .
But this is now a contradiction of the parallel
postulate :
PQand PQare two different lines , each goes
through P and each is parallel to
AB.
This contradiction comes about because
we assumed that APQ  PAB.
So these angles must be congruent.
THEOREM
Let ABC be any triangle. then
A  B  C  180
.
Proof
Let PAQbe the line through A parallel to BC
such that  B and BAPare opposite interior
angles and C and CAQ are opposite
interior angles, as in figure. so B  BAP
and C  CAQ . Hence
A  B  C  A  BAP  CAQ
=180.
Since  A , BAP and CAQ all together
make a straight line.
Let ABCD be any quadrilateral. then
A  B  C  D  360
We draw the diagonal
AC thus breaking the
quadrilateral into two triangles. Note that
A  B  C  D 
CAB  CAD  B  ACB  ACD  D 
(ACD  D  DAC )
+
(CAB  B  BCA).
The first sum of the last expression represents
the sum of the angles of ACD
and the second sum represents the sum of the
angles of CAB. Hence,
each is 180 and together they add up to 360.
COROLLARY(SAA)
ABC and DEFassume that
A  D, B  E and BC  EF.
In
then ABC  DEF.
Given a quadrilateral ABCD, the following
are equivalent:
1. AB CD and AD BC .
2. AB  CDand AD  BC.
3.The diagonals bisect each other.
LEMMA
l
l
Let be a line. P a point not on .
And A and B distinct points on
l such that
PA is perpendicular to l .Then PA PB .
L
THEOREM
l1 and l2 be parallel lines and let P and Q
be points on l 2 Then the distance from P to l
1
equals the distance from Q to l1
Let
Proof
Draw lines from P and from Q perpendicular
to
l1, Meeting l1at B and at C , respectively.
Since
PBC  90
and
QCB  90,
these angles are congruent , Moreover
.
QBC is congruent to the supplement ofPBC.
l2
l1
So
PB QC By opposite interior angles.
Similarly , PQ BC. Therefore PBCQ
must be a parallelogram, since opposite
sides are Parallel . Hence
PB  QC
, as claimed.