Let $H$ be th foot of altitude from $A$. Clearly, $H,M,K,N$ are concyclic (nine-point circle) and $\frac{XK}{YN}=\frac{AB}{AC}=\frac{HK}{HN}$ and so, $H$ is the Miquel point of the quadrilateral $XKNY$. Now, we get points $H,M,X,Y$ concyclic and let $Z'$ be the anti-podal point of $M$ in the circle $(HMXY)$. We get that $Z'X \perp XK$ and $Z'Y \perp YN$ and so, $Z'=Z$ and we have $ZH \perp BC$. Thus, result follows.

Lemma: $X$, $A$ ,$Y$ are collinear. $\odot (AXB)$ $\cap$ $BC$ = $H$, so $H$ $\in$ $\odot (AYC)$. It´s easy to see: $\angle$ $YXM$ = $\angle$ $XYM$ ( $\angle$ $ AKM$ = $\angle$ $ ANM$ ),so $XY$ $ \perp $ $ZM$ $\longrightarrow$ $XZ$ = $YZ$. Then $Z$ it´s in the radical axis of $\odot (AXB)$ and $\odot (AYC)$, so $Z$,$A$ and $H$ are collinear,done. Proof of lemma: It´s easy to see: $2$ $\angle$ $AXM$ = $\angle$ $ANM$ = $2$ $\theta$ ,so $\angle$ $YAN$ = $\theta$ . Then $X$,$A$ and $Y$ are collinear.

A generalizations of this problem: Let $\triangle ABC$. Construct two semicircles with diameter $AB,AC$ outside of $\triangle ABC$. Let $ P$ is any point on side $ BC$ ($ P$ is not coincide with $ B$ or $ C$ ), through $ P$, construct $ PX \parallel AC$ and $ DF \parallel AB$ with $ X$ and $ Y$ on the two semi-circles respectively. The perpendicular drawn from $X$ to $PX$ and perpendicular drawn from $Y$ to $PY$ intersect at $Z$. Prove that $AZ \perp BC$. Proof. Let $AH \perp BC$ at $H$, $M$ is midpoint of $AP$. From Cyclic Pointswe have $A, P, X, Y$ are concyclic. Let $\odot (N)\equiv (APXY)$ then $MN$ is perpendicular bisector of $HM\Rightarrow AZ\parallel MN\perp BC$. DONE.

Note by extraversion that $X,Y$ lie on the external angle bisector of $\angle BAC$, and that $MX=MY$; letting $H$ be the foot of the altitude from $A$, we claim that $M, H, X, Y$ are concyclic. Define $L\equiv XY\cap BC$; it suffices to show that $MH\cdot ML=MX^2$. WLOG $AB<AC$; compute $$MH\cdot ML=\bigg(\frac{a}{2}-c\cdot\frac{a^2-b^2+c^2}{2ac}\bigg)\bigg(\frac{ac}{b-c}+\frac{a}{2}\bigg)=\frac{(b+c)^2}{4}$$On the other hand, $$MX^2=(MK+KX)^2=\bigg(\frac{b}{2}+\frac{c}{2}\bigg)^2$$so we get the conclusion. Then $\angle AHM=\frac{\pi}{2}=\angle ZXM=\angle ZHM$, as desired.

Dear Mathlinkers, 1. according to Reim and pappus theorem, , A and Y are collinear 2. according to the pivot theorem, the circle with diameter ZM pass throug the foot of the A-altitude and we are done... Sincerely Jean-Louis

Let $\omega_1$ be the circle with diameter $AB$ and $\omega_2$ be the circle with diameter $AC$. Now let $\omega_1\cap BC=K$, hence $\angle AKB=90^\circ$ and $\omega_2\cap BC=K'\implies \angle AK'C=\angle AK'B=90^\circ\implies K\equiv K'$. Now $AZ$ must be the radical axis of $\omega_1$ and $\omega_2$. Hence, $AZ\perp KN\implies AZ\perp BC$ as $KN\perp BC$.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%205.pdf p. 46-47. Sincerely Jean-Louis

It is easy to see that $ZYMX$ is cyclic with diameter $ZM$. Furthermore, $MX=MK+KX=MY=MN+NY=\frac{AB+AC}{2}$. Therefore, we have that $ZX=ZY,$ or that $\text{pow}(Z, \omega_c)=\text{pow}(Z, \omega_b)$. This means that $Z$ is on the radical axis of $\omega_c$ and $\omega_b$. However, it is easy to see that the radical axis of $\omega_c$ and $\omega_b$ is just the $A$-altitude(just note the right angle to the diameter), so we are done.

$\angle X K A=2 \angle Z X A$. Also $\angle Y N A=2 \angle Z Y A$ By mid point theorem, $AC \parallel XM$ , so $\angle X K A$=$\angle Y N A$. So, $\angle Z X A$=$\angle Z Y A$. So, $XZ=ZY$. So, $Z$ lies on the radical axis of the two circles (can be proved by power of point). By mid point theorem, $KN \parallel BC$. So, $AZ \perp BC$ (Hence Proved)

Let $H$ the foot of altitude from $A$ to $BC$. We need to prove that $Z,A,H$ are collinear. Lemma 1. $X,A,Y$ are collinear Proof: It´s enough to prove that: $\angle{XAK}+\angle{CAB}+\angle{YAN}=180$ $\Leftrightarrow$ $\frac{180-\angle{XKA}}{2} + \angle{CAB} + \frac{180-\angle{YNA}}{2}=180 $ $\Leftrightarrow$ $180-\angle{XKA}+ 2\angle{CAB} + 180-\angle{YNA} = 360$ $\Leftrightarrow$ $\angle{XKA}+\angle{YNA}=2\angle{CAB}$ But, $\angle{XKA}=\angle{YNA}=\angle{CAB}$. So, it´s proved. Lemma 2. $ZX=ZY$ Proof: We know that $\angle{XKA}=\angle{ANY}$ $\Rightarrow$ $\angle{AXK}=\angle{AYN}=\frac{180-\angle{XKA}}{2}$ $\Rightarrow$ $90-\angle{AXK}=90-\angle{AYN}$ $\Rightarrow$ $\angle{ZXA}=\angle{ZYA}$. q.e.d Then, since $ZX=ZY$, the power of $Z$ to both circles are the same. So, $Z$ it´s in the radical axis of $(XAB)$ and $(AYC)$. But $H \in (XAB)$ and $H \in (AYC)$. Then, $A,Z,H$ are colilinear.

Let AH be perpendicular to BC. we will prove H,A,Z are collinear. step1 : X,A,Y are collinear. ∠XAK = ∠XKB/2 ---> ∠BKM = ∠BAC = 180 - 2∠XAK ∠YAN = ∠YNC/2 ---> ∠CNM = ∠CAB = 180 - 2∠YAN so ∠XAK = ∠YAN and ∠XAK + ∠KAN + ∠NAY = 180 so X,A,Y are collinear. step2 : XZYMH is cyclic. first note BXAH and CYAH are cyclic. ∠MXY = ∠XAK = ∠NAY = ∠CHY = ∠MHY ---> XYMH is cyclic. ∠MXZ = 90 = ∠MYZ ---> XZYM is cyclic. so XZYMH is cyclic. now we have ∠ZHM = ∠ZXM = 90 and we had ∠AHM = 90 so H,A,Z are collinear as wanted. we're Done.

Let $AB\cap YZ=K$ and $AC\cap XZ=L$ $\angle MAX=90-\ \frac{\angle A}{2}=\angle YAN \implies \angle MAX+\angle YAN+\angle CAB=180$ $\implies X,A,Y$ are collinear. $\angle ZXY=90-\angle AXK=\frac{\angle A}{2}$ and $\angle XAL=\angle YAC=90-\frac{\angle A}{2} \implies CA\perp XZ$ and similarily $AB \perp YZ$ So $XLYC$ and $BXKY$ are cyclic. \[AL.AC=AX.AY=AK.AB\]$\implies KLCB$ are cyclic. Also we know that $ZLKA$ are cyclic. $\angle AZL=\angle AKL=\angle BKL=\angle BCL=\angle C$ $\angle BAZ=180-(\angle C+\frac{\angle A}{2}-90+\frac{\angle A}{2})=180-\angle C-\angle A+90=\angle B+90$ which gives the result.