Bayes rule with two events

Question

Could someone explain this step?

$$ \begin{equation} \label{eq:bayes} P(A|B,C) = \frac{P(B |A,C)P(A|C)}{P(B|C)} \end{equation} $$

Answer 1

Multiplying both sides with $P(B\mid C)P(C)$ we find:

$$P\left(A\mid B,C\right)P\left(B\mid C\right)P\left(C\right)=P\left(A\mid B,C\right)P\left(B,C\right)=P\left(A,B,C\right)\tag1$$ and: $$\frac{P(B|A,C)P(A|C)}{P(B|C)}P\left(B\mid C\right)P\left(C\right)=P(B|A,C)P(A|C)P\left(C\right)=$$$$P(B|A,C)P\left(A,C\right)=P\left(A,B,C\right)\tag2$$

Observe that the RHS of $(1)$ is the same as the RHS of $(2)$.