Proving $B$, $C$, $D'$ and $E'$ to be concyclic iff $AB+AC=3BC$? 
Let $ABC$ be a triangle with incenter $I$. The incircle of $ABC$ touches $AC$ at $D$ and $AB$ at $E$. Let $DD'$ and $EE'$ be the diameters of the incircle. Prove that $B$, $C$, $D'$ and $E'$ are concyclic if and only if $AB+AC=3BC$.

The 'only if' condition was not in my problem, but experimentation suggests is true as well, so I added it here.

I drew this on GeoGebra and found that $I$ was also on this circle. I did some work, and was able to prove that if $H$ is the circumcenter of $BIC$ then $HD'=HE'$ holds for every triangle, using congruence. However, I am totally unable to make appropriate use of the condition $AB+AC=3BC$. Can anyone help? :)
 A: I'm assuming that the goal here is to show

All five points $I$, $B$, $C$, $D^\prime$ $E^\prime$ are concyclic if and only if $|\overline{AB}| + |\overline{AC}| = 3|\overline{BC}|$.

(If we were only to show that $I$, $B$, $C$, $D^\prime$ are concyclic, then, by symmetry of argument, $E^\prime$ would have to lie on the common circle, and vice-versa. Likewise, $I$, $B$, $D^\prime$, $E^\prime$ are concyclic if and only if $C$ is on the common circle. Thus, so long as $I$ is one of the must-have points, we might as well be talking about all five points at once.)
We begin with a construction that has nothing to do with the length condition.
Take $B^{\prime\prime}$ and $C^{\prime\prime}$ on $\overleftrightarrow{BC}$ as shown, such that $\overline{BB^{\prime\prime}} \cong \overline{AB}$ and $\overline{CC^{\prime\prime}} \cong \overline{AC}$. Then $\triangle ABB^{\prime\prime}$ and $\triangle ACC^{\prime\prime}$ are isosceles, with $\overline{BB^\prime}$ and $\overline{CC^\prime}$ the bisectors of their vertex angles (so that they each contain $I$) and perpendicular bisectors of their bases (so that $B^\prime$ and $C^\prime$ are midpoints).

We glean two key facts:


*

*Because $\angle IDA$, $\angle IEA$, $\angle IB^\prime A$, $\angle IC^\prime A$  are all right angles, Thales' Theorem guarantees that they lie on the circle with diameter $\overline{IA}$. Moreover, $\bigcirc D^\prime I E^\prime$ is the reflection of that circle in $I$, and it will contain $B$ and $C$ if and only if those points are the reflections of $B^\prime$ and $C^\prime$; that is, if and only if $\overline{IB}\cong\overline{IB^\prime}$ and $\overline{IC}\cong\overline{IC^\prime}$.

*$\overline{B^\prime C^\prime}$ is a "midpoint segment" of $\triangle AB^{\prime\prime}C^{\prime\prime}$, so that $\triangle BIC \sim\triangle B^\prime I C^\prime$ and $$2|\overline{B^\prime C^\prime}| = |\overline{B^{\prime\prime}C^{\prime\prime}}| = |\overline{BB^{\prime\prime}}| + |\overline{CC^{\prime\prime}}| - |\overline{BC}|$$ so that $$|\overline{AB}|+|\overline{AC}| = |\overline{BC}| + 2 |\overline{B^\prime C^\prime}|$$ The right-hand side equals $3|\overline{BC}|$ if and only if $\overline{BC}\cong\overline{B^\prime C^\prime}$; that is, if and only if $\overline{IB}\cong\overline{IB^\prime}$ and $\overline{IC}\cong\overline{IC^\prime}$.
The same congruences are necessary and sufficient for both (1) and (2), making those circumstances logically equivalent. $\square$
