Solution to a-concentric-circle-problem This question has been solved via trigonometry. I am trying to solve it geometrically.
Referring to the figure, I translate DE to BB’ (through //gms $DEE’D’$ and $BE’D’B’$). Also, I rotate $AB$ to $AA’$. As a result, the condition $DE + BC = AB + AC$ is equivalent to saying $\triangle CA’B’$ is isosceles with $CA’ = CB’$.

Let the incenter of $\triangle ABC$ be I. Then, CI will perpendicularly bisect $A’B’$ at $C_1$. It should be clear that all the same color marked angles are equal.
The altitude AH and various angle bisectors through A will divide $\angle BAC$ into 5 sections. Their sizes are expressed in terms of x and y as shown. Then, $\angle DAF = \angle HAE = \angle EAC = x + y$.
It should also be clear that APIQ is cyclic with AI as a diameter. To prove that I is also the circumcenter of $\triangle ADE$, it suffices to show I is the orthocenter of $\triangle AUV$. 
There are at least three ways to achieve that:- (1) If we can prove that M and N are also con-cyclic with APIQ; or (2) If we can prove that $\triangle CAD$ is isosceles with $CA = CD$; or (3) $AA’ = AB = … = BD + DE = BD + BB’ = DB’$
I tried all those approaches but not getting anywhere. Any idea?

I asked this question, it has been solved via trigonometry as you mentioned but some days ago I solve it with a shorter solution because it is too easy I just summarize it :
it is clear if  $\angle A$$=90$ we are done therefor at first suppose  $\angle A$$>90$ obviously $BE$>$AB$ and $DC$>$AC$ so $DE$>$AB$+$AC$-$BC$ , contradiction.
now suppose $\angle A$$<90$ similarly $DE$<$AB$+$AC$-$BC$ ,contradiction 
hence $\angle A$=90
