Triangle angle bisectors, trisectors, quadrisectors,
Q1:
The "low" angle quadrisectors coming from $B$ and $C$ (i.e. the ones closer to $\overline{BC}$) meet on the angle bisector coming from $A$ iff $ABC$ is isosceles (with $AB = AC$).
Proof:
The difficult proof is the "only if". Let $O$ be the incenter of $ABC$, where the bisectors meet. Then the angle quadrisectors of $ABC$ are the angle bisectors of $OBC$ - so they meet at a point on the angle bisector coming from $O$. Therefore, if they meet on the angle bisector coming from $A$, then $\overline{AO}$ bisects the angle $\angle BOC$. But then by supplementary angles, $\angle AOB \cong \angle AOC$ - and by the definition of angle bisector, $\angle OAB \cong \angle OAC$, and $\overline{OA} = \overline{OA}$, so by ASA, $\triangle AOB \cong \triangle AOC$, and $\overline{AB} \cong \overline{AC}$.