We'll show that the following is true: \[b{k \choose 2}= h{c \choose 2}\]. Each side of the equation counts all the distinct sets of one teacher and two different students that he/she teaches. In the left-hand side, for each of the $b$ teachers, they can be grouped with all possible 2-subsets of the $k$ students they teach. In the right hand-side, all pairs of students amongst all $c$ students can be grouped with any of the $h$ teachers that teach both of them. The two sides are equal since we are counting the same entity. Rearranging gives the desired $\frac{b}{h}= \frac{c(c-1)}{k(k-1)}$.