Consider the right isosceles $\vartriangle ABC$ at $ A$. Let $L$ be the intersection of the bisector of $\angle ACB$ with $AB$ and $K$ the intersection point of $CL$ with the bisector of $BC$. Let $X$ be the point on line $AK$ such that $\angle KCX = 90^o$ and let $Y$ be the point of intersection of $CX$ with the circumcircle of $\vartriangle ABC$. Let $Y'$ the reflection of point $Y$ wrt $BC$. Prove that $B - K -Y'$. Notation: $A-B-C$ means than points $A,B,C$ are collinear in that order i.e. $ B$ lies between $ A$ and $C$.