, 2010

Let $I$ be the incentre of the triangle $ABC$, let $I_A$ be the excentre corresponding to the vertex $A$, and notice that the vertices $B$ and $C$ both lie on the circle of diameter $II_A$. The line $DI_A$ meets again the latter circle at point $K$, and the lines $BC$ and $IK$ meet at point $L$ (unless $AB = AC$ in which case the conclusion is obvious). Notice that the line $BC$ is the radical axis of the circles $BEC$ and $BIC$ to deduce that $LB \cdot LC = LI \cdot LK$. On the other hand, $LI \cdot LK = LD^2$, for $K$ is the perpendicular foot dropped from the right-angled vertex $D$ of the triangle $DIL$. Consequently, the point $L$ is the radical centre of the following three circles: the incircle of the triangle $ABC$, the circle $BEC$, and the circle $BIC$. Since the common tangent at $E$ of the first two circles is their radical axis, it must pass through $L$. It follows that $E$ is the reflection of $D$ across the line $IL$, so the lines $DE$ and $IL$ are perpendicular and we are done.