Geometry with Trigonometry

64 The parallel axiom; Euclidean geometry Ch. 5

Thus the sum of the degree-measures of the wedge-angles of a triangle is equal to
180.
Proof.Letlbe the line through
Awhich is parallel toBC.Ifm
is the line throughBwhich is
parallel toAC, then we cannot
havel‖mas we would then have
l‖m,m‖ACwhichwouldim-
plyl‖AC; we would then have
BC‖l,l‖ACand soBC‖AC;
thus asBC∩AC=0 we would/
haveBC=AC; this would make
A,B,Ccollinear and contradict
our assumption.

A

B C

Q l R

m

x

x

y

y

Figure 5.4. Angles of a triangle.

Thusmmeetslat some point,Qsay. Then[A,C,B,Q]is a parallelogram and
[A,B],[Q,C]meet at a pointT.NowQis on the opposite side ofABfromC,so
that∠CBAand∠BAQare alternate angles and so|∠CBA|◦=|∠BAQ|◦. Moreover
[A,B⊂IR(|CAQ)and so|∠CAB|◦+|∠BAQ|◦=|∠CAQ|◦.
ChooseR=Aso thatA∈[Q,R].ThenR∈landRis on the opposite side of
AC fromQ.ButBQ‖ACsoBandQareonthesamesideofAC, and henceB
andRare on opposite sides ofAC.Then∠BCAand∠CARare alternate angles, so
|∠BCA|◦=|∠CAR|◦. Thus

(|∠CAB|◦+|∠CBA|◦)+|∠BCA|◦=(|∠CAB|◦+|∠BAQ|◦)+|∠BCA|◦

=|∠CAQ|◦+|∠CAR|◦

= 180.

COROLLARY.If the points A,B,C are non-collinear, and D=C is chosen so
that C∈[B,D],then|∠ACD|◦=|∠BAC|◦+|∠CBA|◦. Thus the degree-measure of an
exterior wedge-angle of a triangle is equal to the sum of the degree-measures of the
two remote wedge-angles of the triangle.
Proof. For each of these is equal to 180−|∠ACB|◦,asC∈[B,D].

5.3 Ratio results for triangles

5.3.1 Lines parallel to one side-line of a triangle

Let A,B,C be non-collinear points, and with l=AB,m=AC, let≤l,≤mbe natural
orders such that A≤lB,A≤mC. Let D 1 ,D 2 ,D 3 be points of AB such that A≤lD 1 ≤l
D 2 ≤lD 3 ≤lB and|D 1 ,D 2 |=|D 2 ,D 3 |, so that D 2 is the mid-point of D 1 and D 3 .Then
the lines through D 1 ,D 2 and D 3 which are all parallel to BC, will meet AC in points
E 1 ,E 2 ,E 3 , respectively, such that A≤mE 1 ≤mE 2 ≤mE 3 ≤mCand|E 1 ,E 2 |=|E 2 ,E 3 |.