Sec. 10.6 Isometries as compositions 159
LetF=([O,I,[O,J)andF 1 =([Z 0 ,Z 1 ,[Z 0 ,Z 2 )be frames of reference. Let
tO,Z 0 (I)=I 0 ,tO,Z 0 (J)=J 0 ,F′=([Z 0 ,I 0 ,[Z 0 ,J 0 )andα=∠F′I 0 Z 0 Z 1 .Then there
is a unique isometry g such that
g([O,I)=[Z 0 ,Z 1 , g([O,J)=[Z 0 ,Z 2.WhenFZ 1 Z 0 Z 2 is a wedge-angle and so is a right-angle (^90) F′,
g=rα;Z 0 ◦tO,Z 0.
WhenFZ 1 Z 0 Z 2 is a reflex-angle (^270) F′and so its co-supported angle is a right-
angle,
g=s (^12) α;Z 0 ◦tO,Z 0.
Proof. Without loss of generality we take|O,I|=|O,J|=|Z 0 ,Z 1 |=|Z 0 ,Z 2 |=1.
LetZ 0 ∼Fz 0 ,Z 1 ∼Fz 1 ,Z 2 ∼Fz 2 and note thatI∼F 1 ,J∼Fı,z 1 −z 0 =cisα.As
Z 0 Z 1 ⊥Z 0 Z 2 , by 10.1.1(vi) we have
z 2 −z 0 =ı(z 1 −z 0 )whenFZ 1 Z 0 Z 2 is a wedge-angle, (10.6.1)
and
z 2 −z 0 =−ı(z 1 −z 0 )whenFZ 1 Z 0 Z 2 is a reflex-angle. (10.6.2)
In case (10.6.1) we take the transformationZ′=g(Z)where
z′=z 0 +zcisα=z 0 +(z+z 0 −z 0 )cisα.
Then forz=t≥ 0 ,z′=z 0 +t(z 1 −z 0 )sog([O,I)=[Z 0 ,Z 1. Similarly forz=ıt(t≥
0 ),z′=z 0 +t(z 2 −z 0 )sog([O,J)=[Z 0 ,Z 2.
In case (10.6.2) we take the transformationZ′=g(Z)where
z′=z 0 +z ̄cisα=z 0 +(z+z 0 −z 0 )cisα.
Then forz=t≥ 0 ,z′=z 0 +t(z 1 −z 0 )sog([O,I)=[Z 0 ,Z 1. Similarly forz=ıt(t≥
0 ),z′=z 0 +t(z 2 −z 0 )sog([O,J)=[Z 0 ,Z 2.
This establishes the existence ofg. As to uniqueness, suppose thatfis also an
isometry such thatf(F)=F 1. Then by 8.2.1(xii), ifZ∼Fzwe have f(Z)∼F 1
z,g(Z)∼F 1 z,and sof(Z)=g(Z)for allZ∈Π.
COROLLARY.Let f be any isometry. Then f can be expressed in one or other of
the forms
(a)f=rα;Z 0 ◦tO,Z 0 ,(b)f=s (^12) α;Z 0 ◦tO,Z 0.
Proof. In the theorem, takeZ 0 =f(O),Z 1 =f(I),Z 2 =f(J)and consequentlyf
is equal to the functiongas defined in the proof.