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?
 A: 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
A: Most of the suggested angle chasing parts in my post are found to be time wasting. The main point is finding the right lines to construct.
As mentioned in the post, after (1) translating $DE$ to $BB’$; and (2) rotating $AB$ to $AA’$, we get an isosceles $\triangle CA’B’$ with $CA’ = CB’$. Let I be the in-center of $\triangle ABC$.

Let BI, the angle bisector of $\angle ABC$, cut AD at U. From $2b + 2x = 90^0$ (cf $\triangle ABH$), we have $45^0 = b + x = \angle AUI$. Similarly, $\angle AVI = 45^0$; where $V$ is similarly defined as $U$.
Clearly, APIQ is cyclic with AI as a diameter. Let that circle cut AD at M. Therefore, $\angle AMI = 90^0$. Then, $\angle MIU= … = 45^0$. Similarly, $\angle NIV= 45^0$. This further means $CVIM$ and $BUIN$ are straight lines. Then, $CVIM$ is the perpendicular bisector of the chord $AD$ of the green circle. Similarly, $BUIN$ is the perpendicular bisector of the chord $AE$. Result follows.
A: The following geometric proof expands upon the useful suggestion of @math enthusiastic.
First, let $\angle BAC$ be right. Then since $\angle BAE$ is the complement of $\angle EAC$, and $\angle BEA$ is the complement of $\angle EAH$, while$$\angle EAC=\angle EAH$$therefore $$\angle BAE=\angle BEA$$making $\triangle ABE$ isosceles, with $BA=BE$.
The same argument proves that $\triangle CAD$ is isosceles, with $CA=CD$. 
Therefore, since$$BE+CD=BC+DE$$by substitution$$AB+AC=BC+DE$$

Conversely, if it is given that $$AB+AC=BC+DE$$ then $\angle BAC$ is right.  For with $AB$ and $AC$ unchanged, but increasing $\angle BAC$, then $BC$ and $DE$ will increase and$$BC+DE>AB+AC$$which contradicts the supposition. Similarly, diminishing $\angle BAC$ from right to acute, then $BC$ and $DE$ are diminished and$$BC+DE<AB+AC$$again a contradiction.
If follows that under the given conditions $\angle BAC$ is right, and that triangles $ABE$ and $ACD$ are isosceles.
Now draw $BJ$ and $CG$ bisecting $\angle ABE$ and $\angle ACD$, and from their intersection $K$ draw $KF$, $KQ$, and $KT$ perpendicular to $AB$, $AC$, and $BC$.
Then it's clear by AAS congruency that $KT=KF$ in triangles $KTB$, $KFB$, and that $KT=KQ$ in triangles $KTC$, $KQC$. Therefore$$KF=KT=KQ$$and $K$ is the incenter of $\triangle ABC$.
Finally, join $KA$, $KD$, and $KE$. Then since $KG$ and $KJ$ are perpendicular bisectors of $AD$ and $AE$, it follows by SAS congruency that $KA=KD$ in triangles $KGA$, $KGD$, and that $KA=KE$ in triangles $KJA$, $KJE$, so that$$KA=KD=KE$$ and $K$ is the circumcenter of $\triangle ADE$.
Therefore, under the given conditions, the incenter of $\triangle ABC$ coincides with the circumcenter of $\triangle ADE$.  
