Why is the denominator in a conditional probability the probability of the conditioning event? Quite a simple or at least short question: Why is $ \frac{P(A \cap B)}{P(B)} $ divided by $ P(B) $ for the conditional probability?
$ P(A | B) = \frac{P(A \cap B)}{P(B)} $
Random image to visualize: 
I actually would like to make use out of the tree display which I can't grasp at all. Why would I divide by $ P(A+) $ when I had to move along this path?
Sorry for the mix-up of $A+$ and $ B $..


A more intuitive way to understand is to know that conditioning on an event with strictly positive probability defines a new probability measure.
So instead of looking at the bigger picture (the original sample space, say $X$), we are now restricting to a possibly much smaller sample space (which is $B$ in your case). The intersection $A \cap B$ will be smaller in probability with $X$ being the sample space, but larger when we take the $B$ to be the whole sample space. So we adjust for restricting to a smaller sample space by enlarging the probability of the intersection $A\cap B$. This normalizing process is done by dividing by $\mathbb{P}(B)$. Which is easy to understand, we are thus calculating the relative area of $A\cap B$ with respect to $B$, instead of $X$.
You can also view $\mathbb{P}(A)$ as a conditional probability by noting that $$ \mathbb{P}(A)=\mathbb{P}(A\cap X) = \dfrac{\mathbb{P}(A\cap X)}{\mathbb{P}(X)},$$ where $\mathbb{P}(X)=1$.
