6. In $\triangle A B C$, the angle bisectors of $\angle A B C$ and $\angle A C B$ intersect sides $A C$ and $A B$ at points $D$ and $E$, respectively, and $\angle B D E=24^{\circ}, \angle C E D=18^{\circ}$. Try to find the degree measures of the three interior angles of $\triangle A B C$.
6. As shown in the figure, construct the symmetric point $F$ of $E$ with respect to $BD$, and the symmetric point $G$ of $D$ with respect to $CE$. Connect $FD$ and $EG$. Let $EG$ intersect $BD$ at point $M$.
Since $BD$ and $CE$ are the angle bisectors of $\angle ABC$ and $\angle ACB$ respectively.
Therefore, $F$ and $G$ are both on $BC$.
Since $\angle BDE = 24^\circ$,
Therefore, $\angle BDF = \angle BDE = 24^\circ$.
Also, since $\angle CED = 18^\circ$, and $CE \perp DG$,
Therefore, $\angle GDF = 90^\circ - (\angle DEC + \angle EDB + \angle BDF)$
$$
= 90^\circ - (18^\circ + 24^\circ + 24^\circ) = 24^\circ
$$

So, $\angle GDF = \angle FDB$.
That is, $DF$ bisects $\angle BDG$.
Also, $\angle GEC = \angle CED = 18^\circ$.
In the 6th question figure,
Therefore, $\angle BMF = \angle BME = \angle MED + \angle EDB = 36^\circ + 24^\circ = 60^\circ$.
So, $\angle FMG = 180^\circ - 2 \angle BMF = 60^\circ$.
Therefore, $\angle FMG = \angle BMF$.
That is, $MF$ bisects $\angle BMG$.
So, point $F$ is the excenter of $\triangle DGM$ on side $DM$.
Therefore, $GF$ is the external angle bisector of $\triangle DGM$.
So, $\angle MGF = \frac{1}{2}(180^\circ - \angle EGD) = 54^\circ$.
Therefore, $\angle MBG = \angle BME - \angle MGF = 6^\circ$.
Thus, $\angle ABC = 12^\circ$.
Also, $\angle BCE = \angle BGE - \angle CEG = 54^\circ - 18^\circ = 36^\circ$.
So, $\angle ACB = 72^\circ$.
Therefore, $\angle BAC = 96^\circ$.
In summary, $\angle ABC = 12^\circ$, $\angle ACB = 72^\circ$, $\angle BAC = 96^\circ$.