Let $ABC$ be an isoceles triangle such that $AB=AC$ and $AB\neq BC$.
let $I$ be the center of the incircle of $ABC$.
Let $D$ be the intersection of $(BI)$ and $(AC)$ , and $E$ the intersection of the prerpendicular line to $(AC)$ in $D$ and the line $(AI)$ .
Prove that the symetrical point of $I$ with respect to $(AC)$ lies on the circumcircle of $BDE$
Here is what i did.
Let $(C)$ be the circumcircle of $BEC$ , it is well known that its center is $E$.
since $(ED)\perp (AC)$. The point F symetric to $C$ with respect to $D$ lies on $(C)$ . From $\angle DCI=\angle ICB = \angle IBC$ , the line $(DC)$ is tangent to $(C)$. Let $J$ be the symetric point of $I$ with respect to $D$ . from the power of point , $$DC.DF=DC^2=DI.DB=DJ.DB$$
the point $J$ also lies on $(C)$. let $I'$ be the symetric pont of $I$ with respect to $(AC)$ . Since $[IJ]$ and $[CF]$ bisect each other , $CFJI$ is a parallelogram .From $\angle I'FC=\angle FIC=\angle FJC$ , we find that $I'$ lies on $(C)$
note that $(AC)$ is the internal bisector of $\angle IDI'$ , this shows that $DE$ is the exetrnal bisector of $\angle IDI'$ as $(DE) \perp (AC)$ and we have $EB=EI'$
i don t know what i should do now
Extend CA to cut the circum-circle of $\triangle BED$ at Y.
Let the circum-circle of $\triangle YII’$ cut CA at X. Since II’ is perpendicularly bisected by YAXC, YX must be the diameter of that circle. Then $\angle XI’Y = 90^0 = \angle YDE$.
Result follows.