SAUDI ARABIAN MATHEMATICAL COMPETITIONS

1) From the power of point to circle, we have $${\mathcal{P}}_{A/(BIF)}=AF\cdot AB\text{ and }{\mathcal{P}}_{A/(CIE)}=AE\cdot AC.$$ But it is easy to see that $AF\cdot AB=AE\cdot AC$ then $A$ belongs to the radical axis of two circles $(BIF)$, $(CIE)$. Hence, $A$, $D$, $I$ are collinear. Denote $J=AH\cap BC$ then $AD\cdot AI=AH\cdot AJ$ which implies that $H$, $D$, $I$, $J$ are concyclic, then $HD\perp AI$. Then by considering the radical axis of three circles $(AEF)$, $(JFEI)$, $(JHDI)$, we can conclude that $BC$, $HD$, $EF$ are concurrent.



2) Using sine law in some triangle, we get $$\frac{DB}{\sin \angle BID}=\frac{BI}{\sin \angle BDI}=\frac{BC}{2\sin \angle ABC}$$ and $$\frac{DC}{\sin \angle CID}=\frac{CI}{\sin \angle CDI}=\frac{BC}{2\sin \angle ACB}$$ Then $$\frac{DB}{DC}=\frac{\sin \angle ACB}{\sin \angle ABC}=\frac{AB}{AC}=\frac{LB}{LC}.$$ This implies that $(ADL)$ is the Apollonius circle of triangle $ABC$. Thus $\frac{KB}{KC}=\frac{AB}{AC}$ or $ABKC$ is the harmonic quadrilateral, then $AK$ is the symmedian of triangle $ABC$. Hence, $AI$, $AK$ are isogonal conjugate in angle $\angle BAC$, also in angle $\angle EAF$. Since $G$, $D\in (AEF)$, we have $GD\parallel EF$. Finally, $AS$ is the external bisector of the angle $\angle BAC$, also of the angle $\angle EAF$ then $T$ is the midpoint of major $\mathrm{arc}EF$. From these, we can conclude that $TD=TG$. $\square$