China TST 2004 Quiz 1 P1 wrote: Using $ AB$ and $ AC$ as diameters, two semi-circles are constructed respectively outside the acute triangle $ ABC$. $ AH \perp BC$ at $ H$, $ D$ is any point on side $ BC$ ($ D$ is not coinside with $ B$ or $ C$ ), through $ D$, construct $ DE \parallel AC$ and $ DF \parallel AB$ with $ E$ and $ F$ on the two semi-circles respectively. Show that $ D$, $ E$, $ F$ and $ H$ are concyclic. Solution: Let $AE$ $\cap$ $\odot (AHC)$ $=$ $F'$ (such that $F'$ lies on the semi-circle). By Reim's Theorem, $BE||CF'$. Hence, Pappus' Theorem on $BEDF'CA$ $\implies$ $AB||DF'$. Thus, $F= F'$. Let $\Psi$ denote $\sqrt {AH_C \cdot AB}$ inversion, where $H_C$ is foot from $C$ to $AB$. We want to show $\Psi (HDEF)$ is cyclic. Note, $A\Psi (B)\Psi (H)\Psi (D)\Psi (C)$ is cyclic. Also, $\Psi (DAF)$ is tangent to $AB$ at $A$. \begin{align*} \angle \Psi (E)\Psi (H)\Psi (D)= 180^{\circ}-\angle \Psi (B)A\Psi (D) = 180^{\circ}-\angle A\Psi (F) \Psi (D) \end{align*}Hence, $\Psi (DEFH)$ is Cyclic, as desired $\qquad \blacksquare$