Répondre :
CK =1/3.CB tu peux placer K
CB (1;-1) => 1/3CB = (1/3;-1/3)
soit K(x;y) => CK(x;y-1) = (1/3;1/3) => x = 1/3 et y = 2/3 donc K(1/3;2/3)
AI (1/4;1/2) et AK (1/3; 2/3)
on a que AI = 3/4.AK
vérifie 1/3.3/4 = 1/4 et 2/3.3/4 = 1/2 OK
CB (1;-1) => 1/3CB = (1/3;-1/3)
soit K(x;y) => CK(x;y-1) = (1/3;1/3) => x = 1/3 et y = 2/3 donc K(1/3;2/3)
AI (1/4;1/2) et AK (1/3; 2/3)
on a que AI = 3/4.AK
vérifie 1/3.3/4 = 1/4 et 2/3.3/4 = 1/2 OK