China Girls' Mathematical Olympiad

Let $M$ and $N$ be the feet of the perpendiculars from $D$ to lines $AB$ and $AC$ respectively. Let $E_0$, $F_0$, $M_0$ and $N_0$ be the feet of the perpendiculars from $E$, $F$, $M$ and $N$ to side $BC$ respectively.



It suffices to show that $E_0{F}_0=M_0{N}_0$. Without loss of generality, we may assume that $E$ lies on line segment $BM$. Then $\angle AED<90^{\circ }$. Since $A$, $E$, $D$, $F$ are concyclic, $\angle DFC=\angle AED<90^{\circ }$, and so $F$ lies on line segment $AN$. It follows that we need only to consider the figure above. It suffices to show that $E_0{M}_0=F_0{N}_0$. Note that $E_0{M}_0=EM\cos \angle B$ and $F_0{N}_0=FN\cos \angle C$. We need only to show that $EM\cos \angle B=FN\cos \angle C$, or $$\frac{EM}{FN}=\frac{\cos \angle C}{\cos \angle B}(1)$$ In the right-angled triangles $AMO$ and $ANO$, $MD=AD\sin \angle DAM=AD\sin \angle OAB$ and $ND=AD\sin \angle DAN=AD\sin \angle OAC$. Substitute these equations into equation (2) and we have $$\frac{EM}{FN}=\frac{\sin \angle OAB}{\sin \angle OAC}(3)$$ Since $O$ is the circumcenter of triangle $ABC$, $\angle AOB=2\angle C$ and $\angle OAB=\angle OBA=90^{\circ }-\angle C$, and so $\sin \angle OAB=\sin (90^{\circ }-\angle C)=\cos \angle C$. Likewise, $\sin \angle OAC=\cos \angle B$. Substitute the last two equations into equation (3) and we get the desired equation (1).