Geometry with Trigonometry

62 The parallel axiom; Euclidean geometry Ch. 5

Proof. ChooseR=Pso thatP∈[T,R].ThenR∈nandBandRare on opposite

sides ofAP,sothat∠BAP,∠APRare alternate angles and so have equal degree-

measures. But∠APRand∠TPSare vertically opposite angles and so have equal

degree-measures. Hence|∠BAP|◦=|∠TPS|◦.

A

B

P

R

Q

n

l

◦

◦

Figure 5.1. Alternate angles.

A

B

P

R

Q

S

T

l

n

◦

◦

Corresponding angles.

We call such angles∠BAP,∠TPScorresponding angles for a transversal.

If lines l,m,n are such that l‖m and m‖n, then l‖n.

Proof.Ifl=n, the result is trivial asl‖l, so supposel=n.Iflis not parallel to

n,thenlandnwill meet at some pointP, and then we will have distinct linesland

n, both containingPand both parallel tom, which gives a contradiction by A 7. Thus

parallelism is a transitive relation. Combined with the properties in 4.2.2 this makes

it an equivalence relation.

If lines are such that l⊥n and l‖m, then m⊥n.

Proof.Aslis perpendicular to

nthey must meet at some point

A.Asl‖m, we cannot havem‖

n, as by transitivity that would

implyl‖n. Thusmmeetsnin

some pointP, and if we choose

Bonl,Qonmon opposite sides

ofn,thenwehave|∠APQ|◦=

|∠PAB|◦as these are alternate

angles for parallel lines. Hence

|∠APQ|◦=90 andm⊥n.

A

B◦

P

l ◦ Q

m

n

Figure 5.2.

5.2 Parallelograms.............................

5.2.1 Parallelogramsandrectangles.....................

Definition. Let pointsA,B,C,Dbe such that no three of them are collinear andAB‖
CD,AD‖BC.LetH 1 be the closed half-plane with edgeABin whichClies; as
CD‖ABthen, by 4.3.2,D∈H 1. Similarly letH 3 be the closed half-plane with edge
BCin whichAlies; asAD‖BC,thenD∈H 3. ThusD∈H 1 ∩H 3 =IR(|ABC)
and so by the cross-bar theorem[A,C]meets[B,D in some pointT, which is unique