Chapter 2

One of the basic axioms of Euclidean geometry says that two points

determine a unique line.

EXISTENCE AND UNIQUENESS

Two lines that don’t intersect are called parallel.

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.

PROBLEMGiven a line and a point P not on ‚

construct a line through P and parallel to.

lll

L

let A be any point on , and draw Then draw a line so that as shown in the figure.

This will be the desired line .

AP

PABQPA PQl

If and 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.

PQ l

Now consider . the exterior angle is equal to the interior angle . But this contradicts the exterior angle theorem, which

states that .

PAC APQPAC

PACQPA PQ l

The proof will be by contradiction .

Hence must be parallel to.

I

Given a line and a point P not on,

there exists a line that contains P

and is parallel to.

l l

l

COROLLARY

Given lines AB PQ

,then is parallel PABQPA AB

PQto .

if

as in the figure,and

THE PARALLEL POSTULATE

If is any line and P is a point not on . l

l

l

parallel to.

then there is no more than one line through P

Opposite Interior Angles Theorem

are opposite interior angles.

Let and be parallel lines withPQ

PABQPA

transversal such that and

QPA PABAP

AB

Then .

The proof will be by contradiction.

then we could

If the theorem

and if PABQPA construct a distinct line QP

PABPABQAP through P such that

. QAP are opposite interior angles, their congruence

implies that ABQP .

Since and

was false ,

and are two different lines , each goesQP PQ

But this is now a contradiction of the parallel postulate :

So these angles must be congruent.

we assumed that

through P and each is parallel to .

This contradiction comes about because

AB

PABAPQ .

THEOREM

Let be any triangle. thenABC

180 CBA .

ProofLet be the line through A parallel to

such that

angles and and are opposite

interior angles, as in figure. so

Hence

PAQ BC

BAPB

B BAPC CAQ

CAQBAPACBA

and are opposite interior

CAQC and .

=180.Since , and CAQBAPA all together

make a straight line.

Let be any quadrilateral. then

360 DCBA

ABCD

We draw the diagonal AC thus breaking the

quadrilateral into two triangles. Note that

DCBA

DACDACBBCADCAB

)( DACDACD ).( BCABCAB

+

The first sum of the last expression represents

the sum of the angles of ACDand the second sum represents the sum of the

angles of CAB . Hence,

each is 180 and together they add up to 360.

COROLLARY(SAA)

In and assume that

EFBC ABC DEF

EBDA , and

DEFABC then

.

.

Given a quadrilateral ABCD, the following are equivalent :

AB.1 AD BCCD

CDAB .2 BCAD

and

and

.

.

3.The diagonals bisect each other.

LEMMA

Let be a line. P a point not on. l lAnd A and B distinct points on lsuch that

PAis perpendicular to l. Then PBPA .

L

THEOREM

Let and be parallel lines and let P and Q

Then the distance from P to

2l1lbe points on 2lequals the distance from Q to

1l

1l

Proof

Draw lines from P and from Q perpendicular

to1l, Meeting

1l at B and at C , respectively.

90PBCSince 90QCBand ,

these angles are congruent . , Moreover

is congruent to the supplement of QBC PBC .

2l

1l

So By opposite interior angles.

must be a parallelogram, since opposite sides are Parallel . Hence

PB

PQ

QC

BCSimilarly, . Therefore PBCQ

QCPB ,as claimed .