Question 1
Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^\circ$. Let the height from $A$ cut its side $BC$ at $D$. Let $I, I_B, I_C$ be the incenters of triangles $ABC, ABD, ACD$ respectively. Let also $E_B, E_C$ be the excenters of $ABC$ with respect to vertices $B$ and $C$ respectively. If $K$ is the point of intersection of the circumcircles of $E_CIB_I$ and $E_BIC_I$, show that $KI$ passes through the midpoint $M$ of side $BC$.
Notation clarification supported by the original diagram: the source writes $B_I,C_I$ where the defined points are $I_B,I_C$. Thus the two circles are $(E_C,I,I_B)$ and $(E_B,I,I_C)$, and $K$ denotes their common point other than $I$, as shown in that diagram. This makes the line $KI$ well-defined.
Notation clarification supported by the original diagram: the source writes $B_I,C_I$ where the defined points are $I_B,I_C$. Thus the two circles are $(E_C,I,I_B)$ and $(E_B,I,I_C)$, and $K$ denotes their common point other than $I$, as shown in that diagram. This makes the line $KI$ well-defined.