To prove $B = C$, we need to show $B \subseteq C$ and $C \subseteq B$.

Step 1: Express B and C using set operations

From the property given in the textbook, for any sets $A$ and $B$, we have $A = (A \cap B) \cup (A - B)$. Applying this to set $B$, we can write:

$$B = (B \cap A) \cup (B - A)$$

Similarly,

$$C = (C \cap A) \cup (C - A)$$

Step 2: Use the given condition $A \cap B = A \cap C$

Since $A \cap B = A \cap C$ (given), and intersection is commutative, we have:

$$B \cap A = C \cap A$$

Step 3: Prove $B - A = C - A$

Let $x \in B - A$. Then $x \in B$ and $x \notin A$.

Since $x \in B$, $x \in A \cup B$.

Given $A \cup B = A \cup C$, so $x \in A \cup C$.

Since $x \notin A$, we must have $x \in C$.

Thus $x \in C$ and $x \notin A$, meaning $x \in C - A$.

Therefore, $B - A \subseteq C - A$.

Similarly, let $y \in C - A$. Then $y \in C$ and $y \notin A$.

Since $y \in C$, $y \in A \cup C$.

Given $A \cup C = A \cup B$, so $y \in A \cup B$.

Since $y \notin A$, we must have $y \in B$.

Thus $y \in B$ and $y \notin A$, meaning $y \in B - A$.

Therefore, $C - A \subseteq B - A$.

Hence, $B - A = C - A$.

Step 4: Conclude $B = C$

$$B = (B \cap A) \cup (B - A) = (C \cap A) \cup (C - A) = C$$

∴ Therefore, $B = C$ is proved.