If $A$ and $B$ are independent events, show that $P(A | B \cap C) = P(A|C)$ If $A$ and $B$ are independent events, show that $P(A | B \cap C) = P(A|C)$.

Here $B$ and $C$ are actually dependent events but I suppose that doesn't matter.
It's false. Here is a counterexample.

Consider $3$ independent tosses of a fair coin.

Let $A$ be the event: the first and second tosses come up the same.

Let $B$ be the event: the first and third tosses come up the same.

Let $C$ be the event: the second and third tosses come up the same.

The three events $A,B,C$ are pairwise independent: $P(A)=P(B)=P(C)=\frac12$, $P(A\cap B)=P(A\cap C)=P(B\cap C)=\frac14=P(A)P(B)=P(A)P(C)=P(B)P(C)$.

$P(A)=P(A|B)=P(A|C)=\frac12$, but $P(A|B\cap C)=1$.
$ P(A \vert B \cap C) = \frac{P(A \cap B \cap C)}{P(B \cap C)} $

And,$ P(A \cap B \cap C)$ can be written as $P(A \cap B \vert C)P(C)$.

But since A and B are independent, we can separate them to get :

$ P(A \cap B \vert C)P(C) = P(A \vert C)P(B \vert C)P(C)$ and substituting $P(B\vert C) = \frac{P(B\cap C)}{P(C)}$ gives the answer.