Даны точки A(a), B(b), C(c), D(d).
$AC = CD = DB = \frac{AB}{3}$
a = 5;
$b = 9\frac{1}{2}$.
Решение:
1) $AB = b - a = 9\frac{1}{2} - 5 = 4\frac{1}{2}$;
2) $AC = CD = DB = \frac{AB}{3} = \frac{4\frac{1}{2}}{3} = \frac{9}{2} * \frac{1}{3} = \frac{3}{2} = 1\frac{1}{2}$;
3) $c = a + \frac{AB}{3} = 5 + 1\frac{1}{2} = 6\frac{1}{2}$;
4) $d = c + \frac{AB}{3} = 6\frac{1}{2} + 1\frac{1}{2} = 7\frac{2}{2} = 8$.
Ответ: $C(6\frac{1}{2})$ и D(8).

Даны точки A(a), B(b), C(c), D(d).
$BC = CD = DA = \frac{BA}{3}$
$a = \frac{1}{3}$;
$b = \frac{2}{9}$.
Решение:
1) $BA = a - b = \frac{1}{3} - \frac{2}{9} = \frac{3}{9} - \frac{2}{9} = \frac{1}{9}$;
2) $BC = CD = DA = \frac{BA}{3} = \frac{\frac{1}{9}}{3} = \frac{1}{9} * \frac{1}{3} = \frac{1}{27}$;
3) $c = b + \frac{BA}{3} = \frac{2}{9} + \frac{1}{27} = \frac{6}{27} + \frac{1}{27} = \frac{7}{27}$;
4) $d = c + \frac{BA}{3} = \frac{7}{27} + \frac{1}{27} = \frac{8}{27}$.
Ответ: $C(\frac{7}{27})$ и D$(\frac{8}{27})$.