Suppose that in $\Delta ABC$ we take a point $D$ on $BC$ such that the incircles of $\Delta ACD$ and $\Delta ABD$ are tangent at a point.Now suppose that points $H$ and $I$ are defined on segments $AC$ and $AB$ in the same manner as $D$ was defined on $BC$ above. How can i prove that lines $AD,BH,CI$ are concurrent ?
My attempt:
I've applied Ceva's Theorem and by that I've got :$$ \cfrac{AH \cdot CD \cdot BI}{HC \cdot DB \cdot AI}=1$$
And after that I haven't really made so much progress,some of the thoughts I had was to prove that $AH=AI$ ,$CD=HC$ and $BI=DB$.
But from that i don't simple know how to proceed,for example if we consider$AI=AH$,we see that we have to prove that $IS=UH$ but how to prove that?Same reasoning goes for the other equalities..
Also if you would adopt my approach how would you continue ?
(I am not asking to post solutions only using this approach,but just any hint)
It's easy to see that $$BD-DC=BT_2-CZ=BP-CV=AB-AC.$$ Since $BD+DC=BC$, we have $$BD=\frac{BC+AB-AC}2,DC=\frac{BC+AC-AB}{2}.$$ By symmetry we see what you are looking for: $$BD=BI, CD=CH,AI=AH.$$
Remark: $D,I,H$ are tangent points of the incircle of $\triangle ABC$ to the respective sides.