To show that $A \cap B = A \cap C$ need not imply $B = C$, we can construct a counter-example using specific sets. This demonstrates that the given condition is insufficient to prove that sets $B$ and $C$ are equal.

Let us consider three sets $A$, $B$, and $C$ defined as follows:

$A = \{1, 2\}, \quad B = \{1, 2, 3\}, \quad C = \{1, 2, 4\}$

We will calculate the intersection for the pairs $(A, B)$ and $(A, C)$ to test the given condition.

Using the definition of intersection, $A \cap B$ is the set of elements belonging to both $A$ and $B$. The elements common to $A$ and $B$ are 1 and 2. Therefore,

$A \cap B = \{1, 2\}$

Similarly, the elements common to $A$ and $C$ are 1 and 2. Thus,

$A \cap C = \{1, 2\}$

Comparing the results, we find that $A \cap B = \{1, 2\}$ and $A \cap C = \{1, 2\}$. Hence, the condition $A \cap B = A \cap C$ is satisfied. Both intersections yield the same set of elements derived from set $A$.

However, examining sets $B$ and $C$ directly, we find that $B = \{1, 2, 3\}$ and $C = \{1, 2, 4\}$. Since the element 3 belongs to $B$ but not $C$, and the element 4 belongs to $C$ but not $B$, the sets are clearly not equal. Therefore, $B \neq C$.

This counter-example proves that the equality $A \cap B = A \cap C$ does not necessarily imply $B = C$.