We want to prove Bayes theorem that is conditioned on multiple variables. We want to prove the following formula. $$ P(A \mid B, C) = \frac{P(B \mid A,C) \cdot P(A \mid C)}{P(B \mid C)} $$
First, we have the following identities that are true that we will use for proving the above formula.
From the definition of conditional probability, we have $$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} $$
From Bayes theorem, we have $$ P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)} $$
Proof:
Using the definition of conditional probability, we have $$ \begin{align} P(A|B,C) &= \frac{P(A \cap B \mid C) }{P(B \mid C)} \ \end{align} $$ Again using the definition of conditional probability, we have $$ P(A \cap B \mid C) = \frac{P(A \cap B \cap C)}{P(C)} $$ Adding to the above equation, $$ \begin{align} P(A|B,C) &= \frac{P(A \cap B \mid C) }{P(B \mid C)} \\ &= \frac{P(A \cap B \cap C)}{P(C) \; P(B \mid C)} \\ &= \frac{P(A \cap B \mid C)}{P(B \mid C)} \\ &= \frac{P(A \cap B \mid C)}{P(A \mid C)} \frac{P(A \mid C)}{P(B \mid C)} \\ &= \frac{P(B \mid A, C) \; P(A \mid C)}{P(B \mid C)} \qquad \qquad \text{□} \\ \end{align} $$