Please, hope there is someone who can teach me english. (I learn the planar geometry in chinese, and I don't know a lot of terminologies and expressions in english...)

Really nice problem. Solution for Part $a$: Lemma: $ABC$ is a triangle whose circumcenter is $O$. $D,E,F$ are the altitudes from $A,B,C$ to $BC,CA,AB$ respectively. $P,Q$ are the antipodes of $E,F$ on circles $(OEA)$ and $(OFA)$ respectively. Show that $PE,FQ,AD$ concur and $PQ\parallel BC$. Proof: First let's show that $APQ\overset{?}{\sim} HCB$. \[\measuredangle QAP=\measuredangle QAB+\measuredangle BAP=90+(90-\measuredangle A)=180-\measuredangle A=\measuredangle CHB\]\[AQ=AF.\frac{\sin AOQ}{\sin FOA}=AF.\cot FOA\]\[AP=AE.\frac{\sin POA}{\sin AOE}=AE.\cot AOE\]Note that $\measuredangle CAO+\measuredangle FEA=(90-\measuredangle B)+\measuredangle B=90$ implies $AO\perp EF$. Let $AO\cap EF=K$. \[\frac{AP}{AQ}=\frac{AE}{AF}.\frac{\cot AOE}{\cot FOA}=\frac{AF}{AE}.\frac{\frac{KO}{KE}}{\frac{KO}{FK}}=\frac{AE}{AF}.\frac{FK}{KE}=\frac{AE}{KE}.\frac{FK}{AF}=\frac{\cos C}{\cos B}=\frac{HB}{HC}\]Thus, $APQ\sim HCB$. \[\measuredangle (PQ,AC)=\measuredangle PQA+\measuredangle QAC=(90-\measuredangle C)+(90-\measuredangle A)=\measuredangle B=\measuredangle FEA=\measuredangle (FE,AC)\]Hence $PQ\parallel BC$. Now, by applying Desargues theorem on $\triangle HEF$ and $\triangle APQ,$ since $HE\cap AP=HE_{\infty},HF\cap AQ=HF_{\infty},PQ\cap EF=EF_{\infty},$ these triangles are perspective which yields the concurrency of $AH,PE,QF$.$\square$ Back to the main problem, apply the lemma on $\triangle HBC$ whose orthocenter is $A$. Then, $KL\parallel EF$ and $KF,LE,AD$ are concurrent. Desargues theorem on $\triangle{ALK}$ and $\triangle{DEF},$ the concurrency gives that $AL\cap DE=V,AK\cap DF=U,KL\cap EF=EF_{\infty}$ are collinear. Hence $UV\parallel EF$ as desired.$\blacksquare$ Solution for Part $b:$ Lemma: $ABC$ is a triangle and its incircle whose circumcenter is $I$, is tangent to $BC,CA,AB$ at $D,E,F$ respectively. Let $S_A,S_B,S_C$ be the Sharky-Devil points of $A,B,C$. $IA$ intersects $EF$ at $X$ and $H$ is the altitude from $I$ to $S_BS_C$. Prove that $H,X,I,S_A$ are concyclic. Proof: Set $EF\cap S_BS_C=J$. $\measuredangle JHI=90=\measuredangle IXJ$ yields $J$ is the antipode of $I$ on $(IXH)$. Under the inversion centered at $I$ with radius $ID, \ S_B^*,S_C^*$ are the altitudes from $E,F$ to $DF,DE$ respectively. Since $S_B^*,S_C^*,E,F$ are cyclic, we get that $S_B,S_C,E,F$ are concyclic. Radical axises of $(AS_AEF),(AS_AS_BS_C),(S_BS_CEF)$ must concur. $S_BS_C\cap EF=J$ thus, $J,S_A,A$ are collinear. \[\measuredangle IAJ=180-\measuredangle AS_AI=90\]Hence $S_A$ lies on the circle with diameter $IJ$ which is $(IHX)$.$\square$ Lemma: $ABC$ is a triangle whose circumcenter and orthocenter are $O,H$ respectively. $D,E,F$ are the altitudes from $A,B,C$ to $BC,CA,AB$ and $S$ is the intersection of the tangents through $B,C$ to $(ABC)$. $T$ is the antipode of $O$ on $(OEF)$. Prove that $S,D,T$ are collinear. Proof: Take the inversion centered at $O$ with radius $OA$. Let $K,L$ be the intersection of tangents through $C,A$ and $A,B$. $S^*$ is the midpoint of $BC$. $D^*$ is the Sharky-Devil point of $S$ on $\triangle SKL$. $T^*$ is the altitude from $O$ to $E^*F^*$ where $E^*,F^*$ are $K,L$ Sharky-Devil points respectively. By our previous lemma, we get that $O,D^*,S^*,T^*$ are concyclic which yields the collinearity of $S,D,T$ as desired.$\square$ We will show that both $M,N$ lie on the nine-point circle of $ABC$. Since $G$ is the circumcenter of $(HBC),$ \[\measuredangle GHB=90-\measuredangle BCH=\measuredangle B=90-\measuredangle HEF\]Hence $GH\perp EF$. Let $E',F'$ be the antipodes of $E,F$ on the nine-point circle. $E'H\cap (DEF)=M',F'H\cap (DEF)=N'$ and $GH\cap EM'=T'$. Claim: $S,D,T'$ are collinear. Proof: Pascal at $F'N'FE'M'E$ gives that $H,N'F\cap M'E,FE'\cap EF'$ are collinear. Since $EF'\parallel FE'$ and $EF'\perp EF, \ EF'\cap FE'=GH_{\infty}$. Thus, the line passing through $H$ and $M'E\cap N'F$ must be perpendicular to $EF$. So $T'$ lies on this line which yields $F,N',T'$ are collinear. Let $N$ be the circumcenter of $(DEF)$. Since $NE=NE'$ and $NO=NH,$ we get that $HE'\parallel OE$. Also $HE'\perp ET'$ thus, $\measuredangle OET'=90$.Similarily $\measuredangle T'FO=90$ hence $T'$ is the antipode of $O$ on $(OEF)$. By the lemma, we get that $S,D,T'$ are collinear.$\square$ We have $T'=T,M'=M,N'=N$ hence $M,N$ lie on the nine-point circle of $(ABC)$ as desired.$\blacksquare$

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