A =(2;3) , B =(8;3) , C = (5;7)
D =(1; -8), E =(-11; -8), F = (-5; -16)
wektor AB = [8-2;3-3] = [6;0]
AB = 6
wektor BC = [5-8;7-3] = [-3;4]
BC =\sqrt{(-3)^2}+4^2 = \sqrt{9+16} = \sqrt{25} = 5
wektor AC = [5-2; 7-3] = [ 3;4]
AC =\sqrt{3^2+4^2} = \sqrt{25} = 5
trójkąt ABC jest równoramienny.
wektor DE = [-11-1;-8-(-8)] = [-12;0]
DE = 12
wektor EF = [-5 -(-11); -16 -(-8)] = [6;-8]
EF =\sqrt{6^2 +(-8)^2} = \sqrt{36+64} = \sqrt{100} = 10
wektor DF = [-5-1;-16 -(-8)] = [-6;-8]
DF =\sqrt{100} = 10
\frac{DE}{AB} = \frac{12}{6} = 2
\frac{EF}{BC} = \frac{10}{5} = 2
\frac{DF}{AC} = \frac{10}{5} = 2
Te trójkąty są podobne.