Team Selection Test for IMO 2011

Let $a=BC$, $b=CA$, $c=AB$, and $u=(a+b+c)/2$. Let $P$ be the point where the line drawn parallel to $AB$ through $I$ meets $BD$. Similarly define $Q$. Then $IP=u-b=AF$, $IQ=u-c=AE$ and $\angle PIQ=\angle BAC=\angle FAE$. Hence the triangles $PIQ$ and $FAE$ are congruent.

Let $P^{\prime }$ be the point on the ray $IQ$ with $I{P}^{\prime }=IP$ and $Q^{\prime }$ be the point on the ray $IP$ with $I{Q}^{\prime }=IQ$. Then $P^{\prime }I{Q}^{\prime }$ and $FAE$ are congruent, and since two pairs of their sides are parallel, so is the third: $FE\parallel Q^{\prime }{P}^{\prime }$.



As $I,P,D,Q$ are concyclic, $\angle DIP+\angle I{Q}^{\prime }{P}^{\prime }=\angle DIP+\angle IQP=\angle DIP+\angle IDP=90^{\circ }$. Hence $ID$ is perpendicular to $P^{\prime }{Q}^{\prime }$ and consequently to $EF$.