Geometry with Trigonometry

Sec. 8.2 Isometries 125

(iv) TakeZ 3 =Z 1 so thatZ 1 ∈[Z 2 ,Z 3 ].ThenZ 1 Z 2 =[Z 1 ,Z 2 ∪[Z 1 ,Z 3. Hence

f(Z 1 Z 2 )=f([Z 1 ,Z 2 )∪f([Z 1 ,Z 3 )

=[f(Z 1 ),f(Z 2 )∪[f(Z 1 ),f(Z 3 )

=f(Z 1 )f(Z 2 ),asf(Z 1 )∈[f(Z 2 ),f(Z 3 )].

Iff(Z)∈f(Z 1 )f(Z 2 ), then by the foregoing there is a pointZ 4 ∈Z 1 Z 2 such that

f(Z 4 )=f(Z)andthenasfis one-oneZ=Z 4 ∈Z 1 Z 2.

(v) For

|Z 2 ,Z 3 |=|f(Z 2 ),f(Z 3 )|,|Z 3 ,Z 1 |=|f(Z 3 ),f(Z 1 )|,|Z 1 ,Z 2 |=|f(Z 1 ),f(Z 2 )|,

so by theSSS-principle, these triangles are congruent in the correspondence

(Z 1 ,Z 2 ,Z 3 )→(f(Z 1 ),f(Z 2 ),f(Z 3 )).

(vi) Suppose thatf(H 1 )is not a subset ofH 3. Then there is someZ 4 ∈H 1
such thatf(Z 4 )∈H 4 ,f(Z 4 )∈f(l).Thenf(Z 3 )andf(Z 4 )are on opposite sides of
f(l), so there is a pointWon bothf(l)and[f(Z 3 ),f(Z 4 )]. By (ii) there is a point
Z∈[Z 3 ,Z 4 ]such thatf(Z)=W, and then by (i) and (iv)Z∈l. But this implies that
Z 4 ∈H 1 and so gives a contradiction. Hencef(H 1 )⊂H 3 and by a similar argument
f(H 2 )⊂H 4.
(vii) Take distinct pointsZ 1 ,Z 2 inΠ.IfW∈f(Z 1 )f(Z 2 ),thenby(iv)f(Z)=W
for someZ∈Z 1 Z 2 and soWis a value off.

Z 1

Z

Z 2

H 1

Figure 8.3.

f(Z 1 )

W

f(Z 2 ) H

3

Suppose then thatW∈f(Z 1 )f(Z 2 )and letH 3 be the closed half-plane with edge
f(Z 1 )f(Z 2 )which containsW.LetH 1 be the closed half-plane with edgeZ 1 Z 2 such
that by (vi)f(H 1 )⊂H 3. Take a pointZ∈H 1 such that
|∠Z 2 Z 1 Z|◦=|∠f(Z 2 )f(Z 1 )W|◦and|Z 1 ,Z|=|f(Z 1 ),W|. Then by the SAS-principle
[Z 1 ,Z 2 ,Z]≡[f(Z 1 ),f(Z 2 ),W], and so by (v)[f(Z 1 ),f(Z 2 ),f(Z)]≡[f(Z 1 ),f(Z 2 ),W].
In particular
|∠f(Z 2 )f(Z 1 )f(Z)|◦=|∠f(Z 2 )f(Z 1 )W|◦.Asf(Z)∈H 3 ,W∈H 3 we then have
f(Z)∈[f(Z 1 ),W. But by the congruence we also have|f(Z 1 ),f(Z)|=|f(Z 1 ),W|.It
follows thatf(Z)=Wand soWis a value off.