6. Connecting the diagonals of the regular pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$ intersects to form a regular pentagon $B_{1} B_{2} B_{3} B_{4} B_{5}$, and connecting the diagonals of the regular pentagon $B_{1} B_{2} B_{3} B_{4} B_{5}$ again intersects to form a regular pentagon $C_{1} C_{2} C_{3} C_{4} C_{5}$ (as shown in Figure 1). Then, among the triangles formed by the line segments in Figure 1, there are ( ) isosceles triangles.
(A) 50
(B) 75
(C) 85
(D) 100
6.C.

For any point $P$, the number of isosceles triangles with $P$ as the "vertex" (the common point of the two equal sides) is denoted as $[P]$, then
$$
\begin{array}{l}
{\left[A_{1}\right]=6\left(\triangle A_{1} A_{2} A_{5}, \triangle A_{1} B_{3} B_{4}, \triangle A_{1} B_{2} B_{5},\right.} \\
\left.\triangle A_{1} A_{3} A_{4}, \triangle A_{1} A_{2} B_{5}, \triangle A_{1} A_{5} B_{2}\right) ; \\
{\left[B_{1}\right]=9\left(\triangle B_{1} A_{3} A_{4}, \triangle B_{1} B_{2} B_{5}, \triangle B_{1} B_{3} B_{4}\right. \text{, }} \\
\triangle B_{1} C_{3} C_{4}, \triangle B_{1} B_{2} C_{5}, \triangle B_{1} C_{2} B_{5}, \triangle B_{1} A_{2} A_{5} \text{, } \\
\left.\triangle B_{1} A_{3} B_{4}, \triangle B_{1} A_{4} B_{3}\right) ; \\
{\left[C_{1}\right]=2\left(\triangle C_{1} B_{3} B_{4}, \triangle C_{1} B_{2} B_{5}\right) .}
\end{array}
$$

Since there are no equilateral triangles in Figure 1, each isosceles triangle has exactly one "vertex".
By symmetry, we have
$$
\left[A_{i}\right]=6,\left[B_{i}\right]=9,\left[C_{i}\right]=2, i=1,2,3,4,5 \text{. }
$$

Therefore, the total number of isosceles triangles is $5 \times(6+9+2)=85$.