we have drawn altitude $AH$ in triangle $ABC$ if bisection of the $\angle HAB$ and $\angle HAC$ meet $BC$ at $D$ and $E$. Further, assume that $DE=AB+AC-BC$.
Prove the circumcenter of triangle $ADE$ and the center of the incircle of triangle $ABC$ are coincident.
my attempt: I assumed segment bisectors of $AB$ and $AE$ meet each other at $O$ and I choose $T$ on $BC$ which $\angle BOT$=90, therefore, we have to prove $\angle TOC$=$\angle DAC$ but I wasn't able to continue to a solution actually, I don't know how to use $DE=AB+AC-BC$.
please share your ideas and solutions even if they're not complete.thanks!
you can also use this picture
Here is an ugly solution using trigonometry.
Denote the length $AH$ by $1$, $\angle DAH, \angle EAH$ by $a, b$. Let $P$ be a point on $AB$ such that $PD\perp AB$. Then $AB=AP+PB=1+DH\tan{2a}=1+\tan{a}\tan{2a}$. Similarly $AC=1+\tan{b}\tan{2b}$. Hence the given equation is equivalent to: $$\tan{a}+\tan{b}=\tan{a}\tan{2a}+\tan{2b}\tan{b}+2-\tan{a}-\tan{b}.$$ From $\tan{2a}=\frac {2\tan{a}}{1-\tan^2{a}}$ and write $x, y$ for $\tan{a}, \tan{b}$, this reduces to $$(2+x+y)(xy+x+y-1)=0.$$ Because $x, y>0$, we have $xy+x+y-1=0$. Hence, $\tan{a+b}=\frac {x+y}{1-xy}=1$ i.e. $\angle BAC=2a+2b=\pi/2$. Notice that now $\triangle ABE, \triangle ACD$ is isosceles with $AE, AD$ as bases. The angle bisectors of $\angle ABC, \angle ACB$ therefore bisect $AE, AD$ perpendicularly. The conclusion follows.