122 Translations; axial symmetries; isometries Ch. 8
(vi)If Z 1 =Z 2 ,Z∈Z 1 Z 2 and W=tZ 1 ,Z 2 (Z),then[Z 1 ,Z 2 ,W,Z]is a parallelogram.
Proof.
(i) Formp(Z 1 ,W)≡(
1
2
(x 1 +u),1
2
(y 1 +v))
,mp(Z 2 ,Z)≡(
1
2
(x 2 +x),1
2
(y 2 +y))
and these are equal if and only ifu=x+x 2 −x 1 ,v=y+y 2 −y 1.
(ii) For ifW 3 =tZ 1 ,Z 2 (Z 3 ),W 4 =tZ 1 ,Z 2 (Z 4 ),then
u 4 −u 3 =x 4 +x 2 −x 1 −(x 3 +x 2 −x 1 )=x 4 −x 3 ,
v 4 −v 3 =y 4 +y 2 −y 1 −(y 3 +y 2 −y 1 )=y 4 −y 3.It follows that|Z 3 ,Z 4 |=|W 3 ,W 4 |.
(iii) By (i) the equationtZ 1 ,Z 2 (Z)=Whas the solution given byx=u+x 1 −x 2 ,y=
v+y 1 −y 2.
(iv) By (ii) and (iii) the equationtZ 1 ,Z 2 (Z)=Whas a unique solution and this is
denoted byZ=tZ− 11 ,Z 2 (W). The correspondence fromΠtoΠgiven byW→Zis the
inverse oftZ 1 ,Z 2 and is a function. As by the proof of (iii)
x=u+x 1 −x 2 ,y=v+y 1 −y 2 ,by (i) this inverse function istZ 2 ,Z 1.
(v) For ifZ 1 =Z 2 , in (i) we haveu=x+x 1 −x 1 =x,v=y+y 1 −y 1 =y.
(vi) We denote byTthe common mid-point mp(Z 1 ,W)=mp(Z 2 ,Z). First we note
thatW=Z,asmp(Z 1 ,Z)=mp(Z 2 ,Z)would implyZ 1 =Z 2 .AsZ∈Z 1 Z 2 ,wehave
T∈Z 1 Z 2 and henceW∈Z 1 Z 2. It follows thatT∈ZWas otherwise we would have
Z 1 ∈ZW,Z 2 ∈ZWand soZ∈Z 1 Z 2. The triangles[Z 1 ,T,Z 2 ],[W,T,Z]are congru-
ent in the correspondence(Z 1 ,T,Z 2 )→(W,T,Z)by the SAS-principle. Hence the
alternate angles∠WZ 1 Z 2 ,∠Z 1 WZhave equal degree-measures and soZ 1 Z 2 ‖WZ.
SimilarlyZ 1 Z‖Z 2 W.
Axial symmetries have the following properties:-
(i)In all cases|sl(Z 3 ),sl(Z 4 )|=|Z 3 ,Z 4 |, so that each axial symmetry preserves all
distances.(ii)Each axial symmetry slhas an inverse function s−l^1 =sl.
Proof.
(i) We note that by 6.6.1,sl(Z 3 )≡(
1
a^2 +b^2[(b^2 −a^2 )x 3 − 2 aby 3 − 2 ac],1
a^2 +b^2[− 2 abx 3 −(b^2 −a^2 )y 3 − 2 bc])
,
sl(Z 4 )≡(
1
a^2 +b^2[(b^2 −a^2 )x 4 − 2 aby 4 − 2 ac],1
a^2 +b^2[− 2 abx 4 −(b^2 −a^2 )y 4 − 2 bc]