Couple things to remember for this proof:
$ P(A|B)= P(A \bigcap B)/P(B) $
so
$ P(A \bigcap B)= P(A|B)*P(B) $
Make sure you use the theorem of total probability, which states:
$ P(A)=P(A|C)P(C)+P(A|C^c)P(C^c) \! $
Try and rearrange what we want to proof so it looks like what we know is true.
