Expanding expression based on conditional probability The following expression (1) is supposedly "trivially" equal to expression (2), and hence the details are not shown in the example. But of course, I can't get it right.
(1) p(A|B,C)*p(B|C)
(2) p(A,B|C) = p(A|C)*p(B|C) (either side of this equation is equivalent to (1), whichever is easier)

I suspect the answer will make me look pretty stupid, but I am stuck. Why/how are they equivalent?
 A: I think I've got it...
1) p(A | B, C) * p(B|C)
2) p(A, B, C) * (1 / p(B, C) ) * p(B|C)
3) p(A, B, C) * (1 / p(B, C) ) * p(B, C) * (1/ p(C) )
4) [ p(A, B, C) * (1 / p(C) ) ]  * ( 1/ p(B, C) ) * p(B, C)
5) p(A, B | C) * p(B, C) * ( 1/ p(B, C) )
6) p(A, B | C)

Please let me know if this looks correct. Thank you. (Does that really seem trivial?)
A: The reason it is deemed trivial is because it is trivial that
$$\Pr(A|B)*\Pr(B) = \Pr(A,B)$$
Now, imagine that we restrict the probability space to $C$. Then the equality holds also when all the events are conditioned on $C$.
A: Given the Bayes theorem , P(A,B,C) = P(A|B,C)P(B,C), we can easily prove the statement in the question with the following steps:
P(A|B,C)*P(B|C)  =  ( P(A,B,C)/P(B,C) ) * P(B|C) 

according to Bayes theorem again, we can state: P(A,B,C) = P(A,B|C)P(C) and P(B,C)=P(B|C)P(C)
therefore,
P(A|B,C)*P(B|C)  =  ( P(A,B|C)P(C)/P(B|C)P(C) ) * P(B|C) 

After simplification of the resulting statement we arrive at:
P(A|B,C)*P(B|C)  =  P(A,B|C)

