Prove That \((A\cap B)\times C=(A\times C)\cap(B\times C)\)

Let \[ (x,y)\in (A\cap B)\times C \]

Then \[ x\in A\cap B \quad \text{and} \quad y\in C \]

So,

\[ x\in A,\quad x\in B \]

and

\[ y\in C \]

Therefore,

\[ (x,y)\in A\times C \]

and

\[ (x,y)\in B\times C \]

Hence,

\[ (x,y)\in (A\times C)\cap(B\times C) \]

Now let \[ (x,y)\in (A\times C)\cap(B\times C) \]

Then \[ (x,y)\in A\times C \] and \[ (x,y)\in B\times C \]

Therefore,

\[ x\in A,\quad x\in B,\quad y\in C \]

So,

\[ x\in A\cap B \]

Hence,

\[ (x,y)\in (A\cap B)\times C \]

Therefore,

\[ \boxed{ (A\cap B)\times C=(A\times C)\cap(B\times C) } \]