I am sorry, but is this the same $H$? HuangZhen wrote: Line $EZ$ and line $DF$ intersect at $H$ HuangZhen wrote: $H$ is the orthocenter of $\varDelta{ABC}$

MathPanda1 wrote: I am sorry, but is this the same $H$? HuangZhen wrote: Line $EZ$ and line $DF$ intersect at $H$ HuangZhen wrote: $H$ is the orthocenter of $\varDelta{ABC}$ Yes,I think so...

However, unless I read the problem wrong, the intersection of line $EZ$ and line $DF$ never coincides with the orthocenter of $ABC$.

I don't know what the original problem is, but I can show that if we let $H$ be the orthocenter of $\triangle{XYZ}$ then the problem is true. HuangZhen wrote: In $\varDelta{ABC}$,the excircle of $A$ is tangent to segment $BC$,line $AB$ and $AC$ at $E,D,F$ respectively.$EZ$ is the diameter of the circle.$B_1$ and $C_1$ are on $DF$, and $BB_1\perp{BC}$,$CC_1\perp{BC}$.Line $ZB_1,ZC_1$ intersect $BC$ at $X,Y$ respectively.Line $EZ$ and line $DF$ intersect at $H$,$ZK$ is perpendicular to $FD$ at $K$.If $H$ is the orthocenter of $\textcolor{red}{\triangle{XYZ}}$, prove that:$H,K,X,Y$ are concyclic. Here is my solution to the problem stated above. Let incircle of $\triangle{ABC}$ intersect $AB,BC,CA$ at $D_2,E_2,F_2$ respectively. Let $DF\cap BC=T$ and $D_2F_2\cap BC=T_2$ By Menelaus's, we have $\frac{BT_2}{T_2C}\times \frac{CF_2}{F_2A}\times \frac{AD_2}{D_2B}=1=\frac{BT_2}{T_2C} \times \frac{CF_2}{BD_2}=\frac{BT_2}{T_2C} \times \frac{DB}{CF}$. Again, by Menelaus's with $\triangle{ABC}$ and transversal $TDF$, we have $\frac{CT}{TB}\times \frac{BD}{DA}\times \frac{AF}{FC}=1=\frac{CT}{TB}\times \frac{DB}{CF}$. So $\frac{BT_2}{T_2C}=\frac{CT}{TB}$, this mean that the midpoint of $TT_2$ is $M$, the midpoint of $BC$. It is well-known that $(B,C;E_2,T_2)=-1$ since $AE_2,CD_2,BF_2$ concurrent. Reflect through $M$ give us $(B,C;E,T)=-1$. Pencil at point at $\infty$ of direction $\perp BC$ give us $(B_1,C_1;H,T)=-1$ and then at point $Z$ give us $(X,Y;E,T)=-1$. Let $XH\perp YZ=P, YH\perp XZ=Q$, it is well-known that if $PQ\cap XY=T_3$ then $(X,Y;E,T_3)=-1$, so $T=T_3$. Consider cyclic quadrilateral $XQPY$, note that $\angle{XQY}=\angle{XPY}$, so the quadrilateral is cyclic with center at $N$, the midpoint of $XY$. By Brocard's Theorem, we get that $TH\perp ZN$. Since $ZK\perp TH$, we get $Z, K, N$ collinear. Let $Z_1$ be the point defined by reflecting $Z$ through $N$, we have $Z_1X\parallel ZY, Z_1Y\parallel ZX$. So $\angle{HXZ_1}=90^{\circ}=\angle{HYZ_1}$, and we also have $\angle{HKZ_1}=90^{\circ}$ Hence $X,H,K,Y,Z_1$ lie on the same circle, in other word, $H,K,X,Y$ are concyclic and we are done.

Assuming $H$ is the orthocenter of $XYZ$, I have a solution that someone should check since it seems too simple. Let $DF$ intersect $BC$ at $P$. Note that $(EP;BC)=-1$, so projecting orthogonally to $BC$ onto $DF$, we have that $(HP;B_1C_1)=-1$. Then, we project from $Z$ onto $BC$ to conclude that $(EP;XY)=-1$. Let $X'$ and $Y'$ be the feet of the $X$ and $Y$ altitudes in $XYZ$. Now, if $H$ is the orthocenter of $XYZ$, then there is an inversion about $H$ swapping $EZ,XX',YY',PK$. Then, it suffices to show that $P$ lies on $X'Y'$, which is obviously true since $(EP;XY)=-1$.

I assumed that: $H$ is orthocenter of $\triangle XYZ$ Let $W=DF\cap BC$, since $(B,C,E,W)=-1$ projecting in $BC$ we obtain $(B_1,C_1,H,P)=-1$. On the other hand let $X_1=XH\cap YZ$ and $Y_1=YH\cap ZX$, since $(B_1,C_1,H,P)=-1$ we get $Z(X,Y,E,W)=-1$, let $W'$ $=$ $X_1Y_1$ $\cap$ $XY$ $\Longrightarrow$ $Z(X,Y,E,W')=-1$ hence $W'=W$, so from $XYX_1Y_1$ and $HKX_1Y_1$ are cyclic with diameters $AY$ and $HZ$ respectively and $W$ $=$ $XY$ $\cap$ $HK$ $\cap$ $ X_1Y_1$ we get $WY.WX$ $=$ $WX_1.WY_1=WH.WK$ $\Longrightarrow$ $WX.WY=WH.WK$ hence $XYHK$ is cyclic.

$\angle BDB_1=\angle CFC_1,\angle BB_1D+\angle CC_1F=180^{\circ} $, Sine Rule tells us that $\frac {BB_1}{BD}=\frac {CC_1}{CF}\implies \frac {BB_1}{BE}=\frac {CC_1}{CE}\implies \frac {BB_1}{B_1H}=\frac {CC_1}{C_1H} $ where last equality follows from $BB_1||EH||CC_1$, it also implies $\frac {B_1X}{ZX}=\frac {BB_1}{ZE},\frac {C_1Y}{ZY}=\frac {CC_1}{ZE} $. Now converse of Ceva's Theorem in $\triangle ZB_1C_1$ for cevians $ZH,B_1Y,C_1X $ gives us that they are concurrent using the already proven ratio equalities at appropriate places. Hence it follows that $(B_1C_1\cap XY,E;X,Y)=-1$. Now let$YH\cap XZ\equiv P, XH\cap YZ\equiv Q $ then $PY\perp XZ,XQ\perp YZ $. Then again $(PQ\cap XY,E;X,Y)=-1$. So $B_1C_1\cap XY\equiv PQ\cap XY\implies HK,PQ,XY $ are concurrent. Now $PQ $ is the radical axis of $\odot (XYPQ) $ and $\odot (ZQKHP)$, where both circles exist due to $\angle XQY=\angle XPY=90^{\circ},\angle ZQH=\angle ZKH=\angle ZPH=90^{\circ} $, $XY $ is the radical axis of $\odot (XHY) $ and $\odot (XYPQ) $, if we let $l $ be the radical axis of $\odot (ZQKHP) $ and $\odot (XHY) $ then by Radical Axis theorem $XY,PQ,l $ are concurrent and $H\in l $, then by the result that $PQ,HK,XY $ are concurrent we conclude that $l\equiv HK $. But $K\in \odot (ZQKHP) $ hence $K $ also lies on $\odot (XHY)\implies H,K,X,Y $ are concyclic. Note: I have also assumed $H $ to be the orthocentre of $\triangle XYZ $. Thanks @ TheDarkIord for pointing this out.

H is the orthocenter of $\triangle {XYZ}$,I'm so sorry for my typing mistake.

There is something I would like to add to the problem: $K $ lies on the median through $Z $ of $\triangle XYZ $ and its reflection in the midpoint of $XY $ lies on $\odot (XYZ) $.

In other words, K is the A-Humpty point, now aka HM point from a MR article.

Use the properties of harmonic points and make an inversion about $H$.

Let $U$ and $V$ be the foot of $X,Y$-altitudes of triangle $XYZ$, and $G=BC \cap DF$. Note that $$-1=A(BC, EG)=(B_1C_1, HG)=(XY, EG),$$so $G, U, V$ are collinear. Then $K, H, U, V$ lies on the circle with diameter $ZH$, so by PoP we have $$GK \cdot GH=GU \cdot GV=GX \cdot GY,$$or $(KHXY)$ is cyclic. We're done.

ZK is the median of triangle XYZ.