$(A\cap B)\cup C = A \cap (B\cup C)$ if and only if $C \subset A$ I have a set identity: $(A \cap B) \cup C = A \cap (B \cup C)$ if and only if $C \subset A$.
I started with Venn diagrams and here is the result:

It is evident that set identity is correct. So I started to prove it algebraic:
1) According to distributive law: $(A \cap B) \cup C = (A \cup C) \cap (B \cup C)$
2) ...
I stuck a little. Because $C$ is a subset of $A$. I thought of pulling out: $(B \cup C)$ but it seems wrong step to me.
How to prove this identity having in mind that $C \subset A$?
Updated
Venn diagram for $C ⊈ A$

 A: It’s a little easier to go the other way: $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$, and you’d like to show that this equals $(A\cap B)\cup C$ if and only if $C\subseteq A$.


*

*Suppose first that $C\subseteq A$; how can you simplify $A\cap C$?  

*Now suppose that $C\nsubseteq A$; then there is some $c\in C\setminus A$. Show that this $c$ is an element of $(A\cap B)\cup C$ but not of $(A\cap B)\cup(A\cap C)$, so that these two sets cannot be equal.


Your Venn diagrams show what happens when $C\subseteq A$, so they’re useful in proving one direction of the desired result: if $C\subseteq A$, then the two sets are equal. To see how you might prove the other direction, i.e., that if $C\nsubseteq A$, then the two sets are not equal, you’d do better to look at a Venn diagram showing $C\nsubseteq A$.
A: Here is a full algebraic proof.  Let's first expand the definitions:
\begin{align}
& (A \cap B) \cup C = A \cap (B \cup C) \\
\equiv & \;\;\;\;\;\text{"set extensionality"} \\
& \langle \forall x :: x \in (A \cap B) \cup C \;\equiv\; x \in A \cap (B \cup C) \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\cap\;$ and $\;\cap\;$, both twice"} \\
& \langle \forall x :: (x \in A \land x \in B) \lor x \in C \;\equiv\; x \in A \land (x \in B \lor x \in C) \rangle \\
\end{align}
Now we have a choice to make: do we distribute $\;\lor\;$ over $\;\land\;$ in the left hand side, or $\;\land\;$ over $\;\lor\;$ in the right hand side?  Since this expression is completely symmetric, we arbitrarily choose to distribute on the left hand side, and continue our logical simplification after that:
\begin{align}
\equiv & \;\;\;\;\;\text{"distribute $\;\lor\;$ over $\;\land\;$ on left hand side"} \\
& \langle \forall x :: (x \in A  \lor x \in C) \land (x \in B \lor x \in C) \;\equiv\; x \in A \land (x \in B \lor x \in C) \rangle \\
\equiv & \;\;\;\;\;\text{"extract common conjunct out of $\;\equiv\;$"} \\
& \langle \forall x :: x \in B \lor x \in C \;\Rightarrow\; (x \in A  \lor x \in C \equiv x \in A \rangle \\
\equiv & \;\;\;\;\;\text{"simplify one way of rewriting $\;\Rightarrow\;$"} \\
& \langle \forall x :: x \in B \lor x \in C \;\Rightarrow\; (x \in C \Rightarrow x \in A) \rangle \\
\equiv & \;\;\;\;\;\text{"combine both antecedents"} \\
& \langle \forall x :: (x \in B \lor x \in C) \land x \in C \;\Rightarrow\; x \in A \rangle \\
\equiv & \;\;\;\;\;\text{"simplify antecedent: use $\;x \in C\;$ in other side of $\;\land\;$"} \\
& \langle \forall x :: x \in C \;\Rightarrow\; x \in A \rangle \\
\end{align}
Now we only have to wrap up:
\begin{align}
\equiv & \;\;\;\;\;\text{"definition of $\;\subseteq\;$"} \\
& C \subseteq A \\
\end{align}
A: By Distributivity, 
$$ (A \cap B) \cup C = (A \cup C) \cap (B \cup C ) $$
Suppose $ C \subset A $, then $A \cup C = A $ by definition. Therefore $ (A \cap B) \cup C = A \cap (B \cup C )$
Conversely, suppose $ (A \cup C) \cap (B \cup C ) = A \cap (B \cup C )$. We show $C \subset A $. But, by the equality, it is obvious that $ A \cup C = A $. In other words, $C $ must be inside $A$
