How to prove that these lines are concurrent? 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 ?
 A: Let the tangent of circle $O$ at $AB$ be $M$, also let $O$'s tangent at $BD$ be $X$.
Let the tangent of circle $O_1$ at $AC$ be $N$, also let $O_1$'s tangent at $DC$ be $Y$.
Then obviously $IX=IM$ and $IY=IN$ as triangles $IXB$ and $IMB$ are congruent and $IY$ part similarly. 
Connect $AI$, since $I$ is the incenter of $ABC$, $AI$ is also an angle bisector. Using the similar congruence over triangles $AIM$ and $AIN$ we have $IM=IN$.
Hence $IX=IY$. But $XD=DY$ as they are both equal to $DG$.
This is enough to show $ID$ is a median of isoceles triangle $IXY$ and hence $ID\perp XY$ or in other words $ID\perp BC$.
Hence $ID$ is the same line as your line $n$. QED.
A: In my solution I will call the tangents of the circles, $Q$ the tangent of the incircle of $\Delta ABD$ to $AB$ and $M$ to $BC$, N the tangent of incircle of $\Delta ADC$ to $BC$ and $S$ to $AC$.
Fistly notice that $AG$ is the radical axis from the two circles, so it is obvious that $AQ^2=AG^2=AS^2$, hence $AQ=AG=AS$.
Let now call $z=BM=BQ$ because they are tangent to the incircle $ABD$, for the same reasoning $y=CN=CS$ and $t=AQ=AS$, $x=DM=DN$ because $D$ is on the common tangent of the two circles and it is on the radical axis.
Therefore $z+2x+y=a$, $z+t=c$, $t+y=b$. 
By summing all the equations we get $2z+2x+2y+2t=a+b+c$, and $z+x+y+t=\cfrac{a+b+c}{2}$. Now observe that $y+t=b$, so by substiting it we get that $z+x=\cfrac{a+c-b}{2}$. So the point D is the point of tangency from the incircle of $\Delta ABC$ and therefore it the incenter of $ABC$ is perpendicular to it.
Q.E.D   
