Question: Let's say we have a problem of predicting whether a storm is coming or not. We have a model `P(storm is coming | how many clouds are outside)`, and have another model `P(storm is coming | how scared the dogs are)`.

My question in general is how to combine these two models in a sensible way. It seems to make sense if I just multiply the two together - however, I don't know what this expression evaluates to:

`P(storm is coming | how many clouds are outside) * P(storm is coming | how scared the dogs are) = ?`

Or, with simpler notation:

`P(A|B)P(A|C) = ?` (where B and C are independent)

~~I have a feeling the answer is `P(A|B)P(A|C) = P(A|B,C)` but I can't prove it.~~ (As many pointed out, this is wrong)

Edit: Sorry for the confusion, the question here is "what is `P(A|B)P(A|C) = ?`"

Answer: 
Not really a simplification (if that is possible at all) but, if you wish a relation between $P(a \vert b)P(a \vert c)$ and $P(a \vert b,c)$, you could say that 

$P(a \vert b)P(a \vert c) = P(a \vert b,c)^2 \frac{P(c \vert b)}{P(c \vert a,b)} \frac{P(b \vert c)}{P(b \vert a,c)}$ 

I think that this relation is more theoretic and fun than useful. In practical cases you would use the simpler

$P(a \vert b) = P(a \vert b,c) \frac{P(c \vert b)}{P(c \vert a,b)}$ 

$P(a \vert c) = P(a \vert b,c) \frac{P(b \vert c)}{P(b \vert a,c)}$

one use is maybe that this tells how you can't give a usefull interpretation for $P(a \vert b)P(a \vert c)$. Say $P(\text{cancer} \vert \text{smoking})=0.1$ and $P(\text{cancer} \vert \text{sports})=0.001$ then how these relate to $P(\text{cancer} \vert \text{smoking and sports})^2$ is dependent on the terms in the fractions and it can be much different from $0.0001$ either larger or smaller and depends on the case.
If the objective is to somehow combine the impacts of B and C on the probability of A, I think it make sense to evaluate the probabilities $P(A|B\cup C) $ and $P(A|B\cap C)$ where:

$P(A|B\cup C) =\frac{P(A|B)P(B)+P(A|C)P(C)-P(A\cap B\cap C)}{P(B\cup C)}$

and
$P(A|B\cap C) =\frac{P(A\cap B \cap C)}{P(B \cap C)}$