To show that \[ A \cap B = A \cap C \] need not imply \[ B = C, \] we give a counterexample.
Let
\[ A = \{1\}, \quad B = \{1, 2\}, \quad C = \{1, 3\}. \]
Now, find the intersection of \( A \) with \( B \):
\[ A \cap B = \{1\} \cap \{1, 2\} = \{1\}. \]
Next, find the intersection of \( A \) with \( C \):
\[ A \cap C = \{1\} \cap \{1, 3\} = \{1\}. \]
Hence,
\[ A \cap B = A \cap C. \]
But clearly,
\[ B = \{1, 2\} \neq \{1, 3\} = C. \]
Therefore,
\[ A \cap B = A \cap C \;\text{need not imply}\; B = C. \]
Hence, the statement is proved by counterexample.