In triangle $ABC$, $G$ is the centroid, $I$ is the incenter, $GI$ || $BC$, what is $\frac{AB+AC}{BC}$?
I have little to no idea what to do with this problem. I drew the diagram and called the point where the angle bisector of $B$ and the median from $C$ intersect $P$, so $\triangle PGI$ is similar to $\triangle PCB$. Additionally, since medians split each other into $2:1$ ratios, I have that $GC:GD$ (where $D$ is the midpoint of $AB$)is $2:1@. But that's about all I have.
I need a bit of help.
The area of the triangle is
$$Area_{ABC} = \frac12 h BC = \frac 12 r (AB+AC+BC)\tag 1$$
where $h$ is the altitude from A to BC and $r$ is the incircle radius.
Since G is the centroid, the distance from G to BC is one-third of $h$, which happens to be $r$ because $I$ and $G$ are at the same height, i.e. $h=3r$. Plug into (1) to obtain $3BC = AB+AC+BC$, or
$$\frac{AB+AC}{BC}=2$$