$\require{cancel}$ Sorry for the grammatical mistake in the title; it was needed to keep the title under 150 characters.
$P$ is a point on a hyperbola whose focal points are $F_1$ and $F_2$. $Q$ is on the line that bisects $\angle F_1PF_2$. Prove $|PF_1-PF_2|>|QF_1-QF_2|$. This means that line $\overline{PQ}$ only intersects the hyperbola in one place - at $P$.
What I've tried is drawing the ellipse which has focal points $F_1$ and $F_2$ which intersects $P$ and drawing the tangent line, but I'm not really sure where to go from here.
I'd prefer a geometric solution over an algebraic, but either are fine.
EDIT: Ignore this, I messed up
EDIT: I've have a solution, but I never used the bisector fact. Can you check it for me?
$QP+PF_1>QF_1$, by triangle inequality. Therefore, $PF_1>QF_1-QP$.
$QP+PF_2>QF_2$, by triangle inequality. Therefore, $PF_2>QF_2-QP$.
Subtracting these two equations, I get:
$$ PF_1-PF_2>QF_1-QP-(QF_2-QP) $$ $$ PF_1-PF_2>QF_1-QF_2-\cancel{QP}+\cancel{QP} $$ $$ PF_1-PF_2>QF_1-QF_2$$
Where do the absolute values come in? And why did I not need to use the fact that $\overline{PQ}$ bisects $\angle F_1PF_2$? Do I add the absolute values in $PF_1>|QF_1-QP|$ because I'm dealing with magnitudes, resulting in $|PF_1-PF_2|>|QF_1-QF_2| $?
EDIT: Changed G to Q
Consider the mirror image $F_1^*$ of $F_1$ with respect to the line that bisects $\widehat{F_1 P F_2}$.
Then obviously $PF_1=PF_1^*$ and $QF_1=QF_1^*$, but $P,F_2,F_1^*$ are collinear, while $Q,F_2,F_1^*$ are not. The triangular inequality now gives your claim.