Show that if $\mathbb{P}(A|B) = \mathbb{P}(A |\Omega \setminus B)$ and $ 0\lt \mathbb{P}(A),\mathbb{P}(B) \lt 1$ that A and B are independent events

Question

I for some reason can't solve this and I have issues visualizing solutions that come with stochastical independence. It seems intuitive for me though, that if $\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{\mathbb{P}(A \cap B^c)}{\mathbb{P}(B^c)}$ then $B$ and $B^c$ must influence A in the same way, which implies they do not influence $A$ at all. This is no proof though, I would appreciate some intuitions/ help.

The definition of independence: $\mathbb{P}(A \cap B) = \mathbb{P}(A)\cdot \mathbb{P}(B)$

$$\mathbb{P}(A|B) = \mathbb{P}(A |\Omega \setminus B) \Leftrightarrow$$

$$\frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{\mathbb{P}(A \cap B^c)}{\mathbb{P}(B^c)}$$

Answer 1

Hint Are you aware of the identity: $$\frac{a}{b}=\frac{d}{c} \,\,\, \text{then} \,\,\,\frac{a}{b}=\frac{d}{c}=\frac{a+d}{b+c}$$ ?

Answer 2

HINT

Suppose that $A$ and $B$ are independent. Then one concludes that: \begin{align*} \mathbb{P}(A\cap B^{c}) = \mathbb{P}(A) - \mathbb{P}(A\cap B) = \mathbb{P}(A) - \mathbb{P}(A)\mathbb{P}(B) = \mathbb{P}(A)(1 - \mathbb{P}(B)) = \mathbb{P}(A)\mathbb{P}(B^{c}) \end{align*}

Can you take it from here?