proving that $OI=DE$, and proving that $OI\perp DE$ In triangle $ABC$  the angle $\angle C= 30^\circ$. If $D$ is a point on $AC$,and $E$ is a point on $BC$ such that $AD=BE=AB$.how  to prove that $OI=DE$, and  how to prove that $OI\perp DE$ where $O$ is the circumcenter, and $I$ is the incenter.
 A: 
1. Consider a circle of radius $|OI|$ with center at $I$. Let $OG || BC$ and $OJ || AC$. Also $|KC|=|KA|$. 
We can write that $|OJ|=2|KF|=2|AK|-2|AF|=|AC|-(|AB|+|AC|-|BC|)=|BC|-|AB| \ \ ^{*)}$. 
As $|AB|=|BE| \Rightarrow |OJ|=|CE|$. Similarly we can prove that $|OG|=|CD|$.
$\angle GOJ =\angle ECD, |OJ|=|CE|, $|OG|=|CD|$ \Rightarrow \triangle GOJ =\triangle ECD $. 
Radius of circimcircle of $\triangle GOJ$ is equal to radius of circimcircle of $\triangle ECD \Rightarrow |HE|=|HD|=|OI|$.
$\angle ECD = 30^\circ \Rightarrow \angle EHD=60^\circ \Rightarrow \triangle EHD$ - equilateral and |DE|=|OI|.

2. $OL$ is a tangent line to a circle at point $O$. Then $\angle GJO =\angle GOL$. 
As $\triangle GOJ =\triangle ECD \Rightarrow$ $\angle CED =\angle GOL$. 
But $GO||BC$ so we have that $OL||DE$
As tangent line $OL\perp OI \Rightarrow DE\perp OI$. And it's true for any $\angle C$.
Both pictures for the case when $|AB|<|BC|$ and $|AB|<|AC|$. Other cases - similarly.
All we need - prove that  $\triangle GOJ =\triangle ECD$.
$^{*)}$ UPD To clarify $|OJ|=2|KF|=2|AK|-2|AF|=|AC|-(|AB|+|AC|-|BC|)=|BC|-|AB|$
a. $OK \perp AB$ as $O$ is circumcenter. Thats why $OKWF$ rectangle and $|OJ|=2|KF|$. (see comments).

b. As I is the incenter $\Rightarrow |AF|=|AM|,|CF|=|CN|,|BM|=|BN| \Rightarrow $
$\Rightarrow |AB|+|AC|+|BC| = 2|AF|+|CF|+|CN|+|BM|+|BN|$. 
$|CF|+|BM|=|CN|+|BN|=|BC| \Rightarrow$  
$\Rightarrow |AB|+|AC|+|BC| = 2|AF|+2|BC| \Rightarrow $
$\Rightarrow 2|AF|=|AB|+|AC|-|BC|$.
It is interesting that points C,I,E,A lies on a circle (and it's true for any $\triangle ABC$). I tried to use this fact to a simpler proof, but I did not succeed. Maybe someone will be lucky.
A: There is not much need of such extravagant constructions. Here is how I have done it:



*

*Note that $\triangle BID \cong \triangle BIC$ by SAS criteria, as we have $BD =BC$ and $\angle BID = \angle BIC$. Similarly, $\triangle CIE \cong \triangle CIB$. Thus, $\angle DIE = 360^{^\circ}- 3\angle BIC=90^{^\circ}- \frac{3\angle A}2=45 ^{\circ}$.

*We know that $\angle BOC= 2\angle A = 60 ^{\circ}$, so $\triangle BOC$ is equilateral and hence $BC=OB$. Thus $\triangle OBD$ is isosceles as well, and noting that $\angle OBA=90^{\circ}-\angle C$, we have $\angle ODB= \frac{90^{\circ}+\angle C}2$. As $\angle IDB= \frac{\angle C}2$, we conclude that $\angle ODI = \angle ODB - \angle IDB=45^{\circ}$ and similarly $\angle OEI = 45^{\circ}$.

*Thus $DO \perp IE$, and $EO \perp DI$, which means that $O$ is the orthocenter of $\triangle IDE$, and thus we have that $IO\perp DE$. Using the fact that $O$ is the orthocenter, we finally also have $ OI= DE\cdot \cot \angle DIE=DE$.
