Why is P(A,B|C)/P(B|C) = P(A|B,C)? I understand $P(A\cap B)/P(B) = P(A|B)$. The conditional is the intersection of A and B divided by the whole area of B. 
But why is $P(A\cap B|C)/P(B|C) = P(A|B \cap C)$?
Can you give some intuition?
Shouldn't it be: $P(A\cap B \cap C)/P(B,C) = P(A|B \cap C)$?
 A: My intuition is the following ...
Conditioning on $C$ means that we are considering only the cases when $C$ is given. Now, suppose that I live in a world where $C$ is always given.
My pepole know nothing about and cannot imagine a world without $C$. For some reason, our mathematicians denote probability of $X$ by $\hat{P}(X)$. They have also already discovered the rule
$$\hat P(A|B) = \frac{\hat P(A\cap B)}{\hat P(B)}\text{.}$$
Now, you as an Earthling, know a world where $C$ is not part of the assumptions in everyday life. So, when you come to our planet you can immediately notice, that every our probability $\hat P(X)$ actually correspond to your $P(X|C)$.
You are immediately able to rewrite the RHS, following the upper discovery:
$$\frac{P(A\cap B\mid C)}{P(B \mid C)}\text{.}$$
But ... What is the LHS? Well, what is the probability of $A$ when $B$ is given when $C$ is (also) given? Precisely $$P(A\mid B\cap C)\text{,}$$
hence the formula.
A: Any probability result that is true for unconditional probability remains true if everything is conditioned on some event.
You know that by definition,
$$P(A\mid B) = \frac{P(A\cap B)}{P(B)}\tag{1}$$ and so if we condition
everything on $C$ having occurred, we get that
$$P(A\mid (B \cap C)) = \frac{P((A\cap B)\mid C)}{P(B\mid C)}\tag{2}$$
which is the result that puzzles and surprises you; you think it should be
$$P(A\mid (B \cap C)) = \frac{P(A\cap B \cap C)}{P(B\cap C)}.$$
So, let's start by setting $D = B\cap C$
write $P(A\mid (B \cap C)) = P(A\mid D)$ as in $(1)$ to get
\begin{align}
P(A\mid (B \cap C)) &= P(A\mid D)\\
&= \frac{P(A\cap D)}{P(D)}\\
&= \frac{P(A\cap (B \cap C))}{P(B\cap C)}\\
&= \frac{P(A\cap B \cap C)}{P(B\cap C)}\tag{3}\end{align}
which is what you think the result should be.  But observe that if you multiply and divide the right side of $(3)$ by $P(C))$, you can get
\begin{align}
P(A\mid (B \cap C)) &= \frac{P(A\cap B \cap C)}{P(B\cap C)}\times \frac{P(C)}{P(C)}\\
&= \dfrac{\dfrac{P(A\cap B \cap C)}{P(C)}}{\dfrac{P(B\cap C)}{P(C)}}\\
&= \dfrac{P(A\cap B \mid C)}{P(B\mid C)}
\end{align}
which is just $(2)$. In short, the intuition about $(2)$ is that it is just $(3)$ (which you agree with) re-written in terms of conditional probabilities conditioned on the same event $C$.
A: Just draw the Venn diagram.  We then have $$\Pr[A \cap B \mid C] = \frac{\text{"1"}}{\text{"C"}}, \quad \Pr[B \mid C] = \frac{\text{"1"} + \text{"2"}}{\text{"C"}}, \quad \Pr[A \mid B \cap C] = \frac{\text{"1"}}{\text{"1"} + \text{"2"}},$$ and the relationship follows by dividing the first expression by the second.

A: \begin{align*}
\frac{P(A,B|C)}{P(B|C)} &= \frac{P(A,B,C)}{P(C)}\frac{P(C)}{P(B,C)} \\
&= \frac{P(A,B,C)}{P(B,C)} \\
&= P(A|B,C)
\end{align*}
