Let $ U $ $ \equiv $ $ XE $ $ \cap $ $ AB, $ $ V $ $ \equiv $ $ ZD $ $ \cap $ $ CA $. From Droz-Farny Theorem we know the midpoints of $ DE, $ $ VX, $ $ UZ $ are collinear, so $ \odot (HDE), $ $ \odot (HVX), $ $ \odot (HUZ) $ are coaxial with common points $ H, $ $ M $. Since $ M $ is the center of spiral similarity of $ \overline{DVZ} $ $ \mapsto $ $ \overline{EXU} $, so $ DV $ $ : $ $ DZ $ $ = $ $ EX $ $ : $ $ EU $ $\Longrightarrow $ there exist a point $ Y $ on $ XZ $ such that $ YD $ $ \parallel $ $ CA $ and $ YE $ $ \parallel $ $ AB $.

Proposed by Ali Golmakani,IRAN

First, by easy angle chasing we can find that $\Delta XCH\sim \Delta HBZ$. now, let $T$ be the foot of altitude from $H$ to $XZ$, then $T$ is the centre of spiral similarity sends $X\to H,\;H\to Z$ then we must have $C\to B$, in other words: $\Delta TXH\sim \Delta TCB\sim \Delta THZ$ so $\angle TXH=\angle TCB,\;\angle TZH=\angle TBC$so the quadrilaterals $TECX,TDBZ$ are both cyclic.$\Rightarrow \angle ETX=C,\;\angle DTZ=B\Rightarrow \angle DTE=A$. now, let $Y'$ be the intersection between $XZ,d_{2}$, we have: $\angle EY'X=\angle BZT=\angle EDT$ so the quadrirateral $DEY'T$ is cyclic$\Rightarrow \angle EY'D=\angle ETD=A$,thus $\angle DY'T=180-\angle EY'X-A=180-\angle AZX-\angle ZAX=\angle AXZ\Leftrightarrow DY'\parallel AC$ so $Y'\equiv Y$, and we are done $\square $

My solution: Let $M=HZ \cap AC$ and $N=HX \cap AB$ $\Longrightarrow$ By Sharygin Geometry Olympiad 2012 - Problem 21 $\Longrightarrow$ $NE.MD=DZ.EX$. Let $Y$ be a point in $XZ$ such that $ YE \parallel AB $ $\Longrightarrow$ $XY.DZ=YZ.MD$ $\Longrightarrow$ $XY.NE=YZ.XE$ since $NE.MD=DZ.EX$ then $ YD \parallel AC $ hence $X,Y,Z$ are collinear...

Nice Problem but It's Obvious, if we have a parabola tangent to three sides then directrix passes through $H$. now $l_1$ & $l_2$ are tangent to parabola, then clearly it remains a simple Brianchon's theorem in $AB, l_1, BC, l_2, AC, \infty$.

Awful solution !!! Assume the parallel to $AB$ through $E$ cuts $XZ$ at $Y'$. Let $P=HZ \cap AC$ and let $Q=HX \cap AB$ Let $\angle BZH= \angle CHX= \beta$. Let $\angle BHZ= \angle CXH= \alpha$. Let $\angle HCB= \phi$ $\Longrightarrow \frac{ZY'}{Y'X} = \frac{QE}{EX}$ Applying the general angle bisector theorem in $\triangle XCX$ and law of sines in $\triangle QEB$ we get: $\frac{ZY`}{Y'X} = \frac{QB \cdot HC ( \sin 90 - \phi )( \sin \phi ) }{ HE \cdot CX ( \sin \beta + \phi )( \sin \alpha + \beta + \phi ) }$ Applying the general angle bisector on $\triangle BZH$ and since $\triangle BZH \sim \triangle CHX$ we get: $1= \frac{DH \cdot ZB ( \sin 90- \phi )}{ZD \cdot BH ( \sin 90 - \alpha - \beta - \phi )} = \frac{HC \cdot ZB ( \sin 90- \phi ) }{ZD \cdot CX ( \sin 90 - \alpha - \beta - \phi ) }$ Dividing both equations we get: $\frac{ZY'}{Y'X} = \frac{ZD \cdot QB ( \sin \phi )( \sin 90 - \alpha - \beta - \phi )}{DH \cdot HE ( \sin \beta + \phi )( \sin \alpha + \beta + \phi )}$ Law of sines in $\triangle DPC$ gives: $1 = \frac{PC ( \sin \alpha + \beta + \phi ) }{DP ( \sin 90 - \phi - \beta ) }$ Multiplying both equations we get: $\frac{ZY'}{Y'X} = \frac{ZD}{DP} \times \frac{PC \cdot QB ( \sin \phi )( \sin 90 - \alpha - \beta - \phi )}{DH \cdot HE ( \sin 90 - \beta - \phi )( \sin \beta + \phi )}$ Applying law of sines in $\triangle BHE$ and $DHC$ and multiplying the result, we get: $\frac{ZY'}{Y'X} = \frac{ZD}{DP} \times \frac{PC \cdot QB}{HC \cdot BH}$ Applying law of sines in $\triangle HPC$ and $\triangle HQB$ and multiplying the result we get: $\frac{ZY'}{Y'X} = \frac{ZD}{DP}$ which completes our proof.

Dear Mathlinkers, for a proof of the Droz-Farny theorem http://forumgeom.fau.edu/FG2004volume4/FG200426.pdf Sincerely Jean-Louis

Let $HZ \cap AC = M$ and $HX \cap AB = N$. Also let $T,Q,R$ be the midpoints of $NZ,DE,MX$, respectively. Finally let $S = XT \cap EY$. Droz-Farny Theorem gives $T,Q,R$ are collinear. And we know that $ES \parallel NT$ so we have $\dfrac{TS}{SX} = \dfrac{NE}{EX}$. Menelaus theorem on triangle $ANX$ with collinear points $B,E,C$ gives $\dfrac{AB}{BN} \cdot \dfrac{NE}{EX} \cdot \dfrac{XC}{CA} = 1$ so we get $$\dfrac{NE}{EX} = \dfrac{AC}{AB} \cdot \dfrac{BN}{XC}.$$Now note that $\angle THQ = \angle THZ + \angle DHQ = \angle TZH + \angle HDE = \angle TZH + \angle BDZ = \angle B$. Similarly $\angle RHQ = \angle C$. Thus we get by the ratio lemma on $\triangle THR$. $$\dfrac{TQ}{QR} = \dfrac{HT}{HR} \cdot \dfrac{\sin(\angle THQ)}{\sin(\angle RHQ)} = \dfrac{NT}{RX} \cdot \dfrac{AC}{AB}.$$Furthermore by the law of sines on $\triangle BQT$ and $\triangle RQC$ we have $$\dfrac{TQ}{\sin(\angle B)} = \dfrac{BT}{\sin(\angle BQT)}$$$$\dfrac{QR}{\sin(\angle C)} = \dfrac{RC}{\sin(\angle BQT)}.$$Dividing this two relationships we get $\dfrac{TQ}{QR} = \dfrac{AC}{AB} \cdot \dfrac{BT}{RC}.$ Recall that we already had $\dfrac{TQ}{QR} = \dfrac{HT}{HR} \cdot \dfrac{AC}{AB}$ so making this two expressions equal we get: $$\dfrac{NT}{RX} = \dfrac{BT}{RC} = \dfrac{NT-BT}{RX-RC} = \dfrac{BN}{XC}.$$Therefore $\dfrac{TQ}{QR} = \dfrac{AC}{AB} \cdot \dfrac{BN}{XC} = \dfrac{NE}{EX} = \dfrac{TS}{SX}$ This gives $QS \parallel RX \parallel DY$ and since $Q$ is the midpoint of $DE$ this means $S$ is the midpoint of $EY$. We can conclude $\dfrac{XE}{ES} = \dfrac{XN}{NT}$ or $\dfrac{XE}{XN} = \dfrac{ES}{NT} = \dfrac{EY}{NZ}$ so $\triangle XEY \sim \triangle XNZ$ which means that $\angle NXY = \angle EXY = \angle NXZ$ so $X,Y,Z$ are collinear as desired.

Let $B'$ and $C'$ be the infinity points along $AC$ and $AB$. Note that the parallel to $HD$ through $B$, the parallel to $AC$ through $E$, and $AH$ concur at the orthocenter of $BHE$. Hence, by Pascal's theorem hexagon $AHDEB'C'$ lies on a conic. But this means that hexagon $HDB'AC'E$ lies on a conic, so by Pascal's theorem the result follows.

Is this I nternational geometry olympiad ??

ABCDE wrote: Note that the parallel to $HD$ through $B$, the parallel to $AC$ through $E$, and $AH$ concur at the orthocenter of $BHE$. Hence, by Pascal's theorem hexagon $AHDEB'C'$ lies on a conic. Why does it follow?

IGO 2015/3 wrote: Let $H$ be the orthocenter of acute triangle $ABC$. Points $D$ and $E$ lie on side $BC$ such that $\angle DHE=90^{\circ}$. Lines $HD$ and $AB$ intersect at point $X$, lines $HE$ and $AC$ intersect at point $Y$. Point $Z$ is defined such that $ZD \parallel AC$ and $ZE \parallel AB$. Prove that $X, Y, Z$ are collinear. Let $P=XD \cap AC$ and $Q=YE \cap AB$. Notice that $$\triangle BHQ \sim \triangle CPH \, \, \text{and} \, \, \triangle BHX \sim \triangle CZH.$$Observe that $Z$ lies on line $XY$ if and only if we have $$(XP, D\infty)=(QY, E\infty).$$Let $R, S$ be points on line $BC$ such that $AR \parallel HD$ and $AS \parallel HE$. Observe $(XP, D\infty) \overset{A}{=} (BC, DR)$ and $(QY, E\infty) \overset{A}{=} (BC, ES)$. Let $F=AH \cap BC$, then the negative inversion about $F$ with power $\sqrt{FH \cdot FA}$ is an involution on line $BC$. Incidentally, $\{B, C\}, \{R, E\}, \{D, S\}$ are pairs of this involution. Thus, $(BC, DR)=(BC, ES)$ and we obtain the desired conclusion. $\blacksquare$

Here's a pretty quick proof of Droz-Farnzy Theorem: Droz-Farnzy Theorem wrote: Let $ABC$ be a triangle with orthocenter $H$. Let $l_1,l_2$ be any two perpendicular lines passing through $H$. Let $(l_1,l_2)$ intersect lines $BC,CA,AB$ at points $(X,X'),(Y,Y'),(Z,Z')$, respectively. Then midpoints of segments $XX',YY',ZZ'$ are collinear. When $\triangle ABC$ is right-angled, the problem is trivial (assume otherwise henceforth). We want centers of $\odot(HXX'),$$\odot(HYY'),\odot(HZZ')$ to be collinear, which is equivalent to the three circles being coaxial. Define $M$ as the Miquel point of lines $BC,CA,AB,l_1$. Clearly $M \ne H$ since $M \in \odot(ABC)$ while $H \notin \odot(ABC)$. Now the reflection of $M$ in lines $BC,CA,AB,l_1$ are collinear on some line $\tau$. Then $H \in \tau$ as $M \in \odot(ABC)$. Since $l_1 \perp l_2$, so reflection of any point in lines $l_1,l_2$ are collinear with $H$. It basically follows that reflection of $M$ in the five lines $BC,CA,AB,l_1,l_2$ (and point $H$) are collinear. So $M$ is the Miquel point of the five lines, which means all of $\odot(HXX')$$,\odot(HYY')$$,\odot(HZZ')$ pass through $M$. So the three circles intersect at another point distinct from $H$, hence they are coaxial. $\blacksquare$

MMP solution? Assume ABC is not right-angled as it is trivial. Fix $\triangle ABC$ and move $D$ along $BC$. $X$ is a projection $\implies degX = 1$. $Y$ is a fixed angle rotation $\implies degY = 1$. $deg D\infty _{AC} = deg E\infty _{AB} = 1 \implies deg Z = 2$ We only need 5 cases. $i) D = B$; $ii) D = C$; $iii) D = AH \cap BC$; $iv) D = \infty _{BC}$; $iv) D = H\infty _{A C} \cap BC$ or $H\infty _{AB} \ \cap BC$.