Geometry with Trigonometry

126 Translations; axial symmetries; isometries Ch. 8

(viii) LetW∈H 3 .Thenby(vii)W=f(Z)for someZ∈Π.IfW∈f(l)then by
(iv)Z∈l⊂H 1 .IfW∈f(l)thenW∈H 4 and by (vi) we cannot haveZ∈H 2 as that
would implyW∈H 4. Thus againZ∈H 1. In both casesWisavaluef(Z)for some
Z∈H 1.
(ix)Bypart(iv)f(l),f(m)are lines. IfZbelongs to bothlandm,thenf(Z)
belongs to bothf(l)andf(m)so these have a point in common.
On the other hand, ifl‖msuppose first thatl=m.Thenf(l)=f(m)and sof(l)‖
f(m). Next suppose thatl=m;thenl∩m=0. We now must have/ f(l)∩f(m)=0,/
as ifWwere on bothf(l)andf(m), by (iv) we would haveW=f(Z)for some
Z∈l,W=f(Z 0 )for someZ 0 ∈m. But by (i)Z=Z 0 so we would haveZon bothl
andm.
(x) By (v) the triangles[Z 1 ,Z 2 ,Z 3 ],[f(Z 1 ),f(Z 2 ),f(Z 3 )]are congruent, and so cor-
responding angles have equal degree-measures.
(xi) Iflandmare perpendicular, letZ 1 be their point of intersection, and letZ 2 ,Z 3
be other points onlandmrespectively. Then as in part (x),∠Z 2 Z 1 Z 3 is a right-angle
and so its image is also a right-angle.
(xii) For any lineland any pointZwe recall thatπl(Z)denotes the foot of the
perpendicular fromZtol. For any pointZ∈Π,letU=πOI(Z)andV=πOJ(Z).
LetO′=f(O),I′=f(I),J′=f(J).ThenO′=I′,O′=J′andO′I′⊥O′J′so that
we can take([O′,I′,[O′,J′)as a frame of reference. LetH 1 ,H 2 be the half-planes
with edgeOI, withJ∈H 1 ,andH 3 ,H 4 the half-planes with edgeOJ, withI∈H 3.
Similarly letH 1 ′,H 2 ′be the half-planes with edgeO′I′, withJ′∈H 1 ′,andH 3 ′,H 4 ′
the half-planes with edgeO′J′, withI′∈H 3 ′.

O I

J

U

V

H 1

H 2

H 4 H 3

Z

O′

J I′

′

Z′

U′

V′

Figure 8.4.

If(x,y)are the Cartesian coordinates ofZrelative to([O,I,[O,J),then

x=

{

|O,U|, if Z∈H 3 ,

−|O,U|, if Z∈H 4.

But ifZ′=f(Z),U′=f(U)we haveU′∈O′I′,andifZ∈OIwe haveZU⊥OIand
henceZ′U′⊥O′I′. It follows thatU′=πO′I′(Z′). Moreoverf(H 3 )=H 3 ′,f(H 4 )=