It is easy to prove $D$ is the point where the $A$-excircle touches the side $BC$. Indeed, if the rays $AB,AC$ touch the $A$-excircle at $P,Q$ respectively and this excircle is tangent to $BC$ at $D'$ then $AP=AQ$. But $BP=BD'$ and $CQ=CD'$ also, so $AB+BD'=AP=AQ=AC+CD'$ so $D=D'$. Consider the homothety that takes the incircle into $A$'s excircle with centre $A$ (the incentre and $A$-excentre both lie on the internal angle bisector of $\angle BAC$). Since $A,X,D$ are collinear and $X,D$ lie on the incentre, excentre respectively, it follows that this homothety takes $X$ to $D$. If $I_1$ is the $A$-excentre then $XI||DI_1$ But $DI_1\perp BC\implies XI\perp BC\implies XI||XE$, which implies $X,I,E$ are collinear. Therefore $XE$ is a diameter of the incircle, meaning $\angle XYE=90^{\circ}$, so $EY\perp AD$. For the second part we'll prove $M$ is the midpoint of $DE$, which is equivalent to proving $BD=EC$. This is well known. Recall that $AB+BD=AC+CD$, meaning that $AB+BD=AC+CD=s$ where $s$ is the semi-perimeter of $\triangle ABC$. Then $BD=s-c$. Let $E,F,G$ be the point of tangency between the incircle and the sides $BC,CA,AB$ respectively. Let $AF=AG=x,BG=BE=y,CE=CF=z$. Then $CF=x=\frac{2x}{2}=\frac{(x+y)+(z+x)-(x+y)}{2}=\frac{a+b-c}{2}=s-c$. Therefore $BD=CE=s-c$. Then $M$ is clearly the midpoint of $ED$, and $I$ is also the midpoint of $EX$, so $MI$ is a mid-line in $\triangle EXD$, meaning $XD=2IM$.

What is the source of this problem? Which country / competition? Thanks.

It's from Hong Kong 2007 I think..