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: 
In my opinion your problem is not about that expression but about modelling. You have different clues (clouds, scared dogs) which provide evidence for a forthcoming event (rain). In other words, if I understand you, your question is actually more about: how to combine different clues?.

This question is dealt with in the field of graphical models. Let me refer to the corresponding [Wikipedia site](https://en.wikipedia.org/wiki/Graphical_model), and the references therein. Specially the paper by David Heckerman ([A Tutorial on Learning With Bayesian Networks](https://www.microsoft.com/en-us/research/publication/a-tutorial-on-learning-with-bayesian-networks/?from=http%3A%2F%2Fresearch.microsoft.com%2Fen-us%2Fum%2Fpeople%2Fheckerman%2Ftutorial.pdf)), and the reference to Christopher Bishop's Machine Learning book.

The example you give is very similar to those presented to describe the explain away effect, that is nicely described in [this video](https://www.youtube.com/watch?v=pyxyYWNo8Qw).
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
$P(A|B\cap C) =\frac{P(A\cap B \cap C)}{P(B \cap C)}$