Point $D$ is chosen on side $BC$ of $\Delta ABC$ such that the incircles of $\Delta ACD$ and $\Delta ABD $ are tangent at $G$. Let line $l$ be the angle bisector of $\angle ABC$ ,line $m$ be the angle bisector of $\angle ACB$,and line $n$ be the perpendicular to $BC$ at point $D$.
Prove that lines $l,m$ and $n$ are concurrent.
My attempt:
I let $ CR$ (angle bisector of $C$) and $DX$ ( line perpendicular to $BC$ at $D$) meet at $I$ ,then ,considering $\Delta CRA$ ,i have to prove that points $B,I,Q$ (where $BQ$ is the angle bisector of $B$) are collinear. For that i've considered $\Delta CRA$ from which i have by Menelaus's Theorem (given that $I,Q,B$ are points that lie on the sides of this triangle)that: $$\cfrac{AB \cdot CQ \cdot RI}{RB \cdot AQ \cdot CI}=1 $$
Applying the Angle Bisector Theorem to $BQ$ i have that $$\frac {CQ}{AQ}=\cfrac{BC}{AB} \tag 1$$ ,doing the same for angle bisector $CR$ i have : $$ BR \cdot AC = AR \cdot BC $$ Multiplying this relation with $(1)$, i have that $\cfrac {AB}{RB}=\cfrac {AB \cdot AC}{BC \cdot AR}$ ,so to prove that the lines are concurrent i have to prove that $\cfrac {RI}{CI}=\cfrac {AR}{AC}$
I am having some trouble proving this,can you give me some hints ?