No 2. A point of K has segment of AB. K isn’t has
AB. And A line of g make g parallel with AB. And distance between K and
AB as twice more than distance between k and g. There is a domain T with range
from AB and the range of g. and then that if p has AB then T (P) = p’ KP 
Answer :
Firstly we must check it. Is it v to v ?
AB 
P 
P
E P F D
A
B
K
g
D’= g(D) F’ P’ E’ = g(E)
a. What
the shape of set P’ . if The P’ move on segment of AB ?
b. Prove
that T is injectif !
c. If
E and F are two point of AB. Are they distance of E’ F’ if E’ = T (E) and F’ =
T(F) ?
Look at the triangle
EKF and E’K F’
For example P 
Every triangle comparation with the
triangle 1 : ½
For ecample the pont of D and E 
D = E is 
D 
The proved of contradiction.
We take DFE get 
That is contradiction when we were using
assumption make a onto function.
