Geometry with Trigonometry

124 Translations; axial symmetries; isometries Ch. 8

(ix)If l and m are intersecting lines, then f(l)and f(m)are intersecting lines. If l

and m are parallel lines, then f(l)and f(m)are parallel lines.

(x)If the points Z 1 ,Z 2 and Z 3 are non-collinear, then

|∠Z 2 Z 1 Z 3 |◦=|∠f(Z 2 )f(Z 1 )f(Z 3 )|◦.

(xi)If l and m are perpendicular lines, then f(l)⊥f(m).

(xii) If a point Z has Cartesian coordinates(x,y)relative to the frame of reference

F=([O,I,[O,J),then f(Z)has Cartesian coordinates(x,y)relative to the

frame of reference([f(O),f(I),[f(O),f(J)).

Proof.
(i) IfZ 1 =Z 2 then|Z 1 ,Z 2 |>0sothat|f(Z 1 ),f(Z 2 )|>0, and sof(Z 1 )=f(Z 2 ).
(ii) IfZ 1 =Z 2 the result is trivial, so suppose thatZ 1 =Z 2. Then for allZ∈[Z 1 ,Z 2 ],
we have|Z 1 ,Z|+|Z,Z 2 |=|Z 1 ,Z 2 |and so|f(Z 1 ),f(Z)|+|f(Z),f(Z 2 )|=|f(Z 1 ),f(Z 2 )|.
It follows by 3.1.2 and 4.3.1 that f(Z)∈[f(Z 1 ),f(Z 2 )] and so f([Z 1 ,Z 2 ])⊂
[f(Z 1 ),f(Z 2 )].

Z 1

Z Z^2

Figure 8.2.

f(Z 1 )

W

f(Z 2 )

Next letW∈[f(Z 1 ),f(Z 2 )].Then|f(Z 1 ),W|≤|f(Z 1 ),f(Z 2 )|=|Z 1 ,Z 2 |. Choose the
pointZ∈[Z 1 ,Z 2 so that|Z 1 ,Z|=|f(Z 1 ),W|;as|Z 1 ,Z|≤|Z 1 ,Z 2 |thenZ∈[Z 1 ,Z 2 ].
Moreover|f(Z 1 ),f(Z)|=|Z 1 ,Z|=|f(Z 1 ),W|.Thenf(Z)andWare both in
[f(Z 1 ),f(Z 2 )and at the same distance fromf(Z 1 )sof(Z)=W. ThusWisavalue
offat some point of[Z 1 ,Z 2 ]. Hencef([Z 1 ,Z 2 ]) = [f(Z 1 ),f(Z 2 )].
(iii) By (i) f(Z 1 )=f(Z 2 ). Suppose thatZ∈[Z 1 ,Z 2. Then eitherZ∈[Z 1 ,Z 2 ]
orZ 2 ∈[Z 1 ,Z]. It follows from part (ii) of the present theorem, that then either
f(Z)∈[f(Z 1 ),f(Z 2 )]or f(Z 2 )∈[f(Z 1 ),f(Z)]. Thus f(Z)∈[f(Z 1 ),f(Z 2 )and so
f([Z 1 ,Z 2 )⊂[f(Z 1 ),f(Z 2 ).
IfW∈[f(Z 1 ),f(Z 2 )chooseZ∈[Z 1 ,Z 2 so that|Z 1 ,Z|=|f(Z 1 ),W|.Thenby
the last paragraphf(Z)∈[f(Z 1 ),f(Z 2 )and as|f(Z 1 ),f(Z)|=|f(Z 1 ),W|,wehave
f(Z)=W. ThusWis a value of fat some point of[Z 1 ,Z 2. Hence f([Z 1 ,Z 2 )=
[f(Z 1 ),f(Z 2 ).