I think the problem is wrong

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tuanh208 wrote: I think the problem is wrong Sorry, I did mistake when translate problem not $AB=CD$ $AB=AD$ And $MN= BM+DN$

let make simmetry of point D wrt AN we get point D' , let line ND' intersect line BC in point M' , easy to see that AD' = AD = AB and angle AD'M = 180 - AD'N = ABM so trangle AD'M = ABM so NM' = BM' + DN easy to see than that point M = M' so easy to see that if we make simmetry of line MN wrt AQ and AP than this new lines intersects in point C on circle so than easy to see that orthocenter of the triangle APQ lies on the segment MN

skytin wrote: ....so than easy to see that orthocenter of the triangle APQ lies on the segment MN can you please explain this last conclusion? how do you get that?

Let $X$ be on $MN$ such that $BM = MX$ and $XN = ND$. Let the circle $O$ centered at $A$ with radius $AB$ intersect $MN$ at $X'$. Then $\angle ABX' + \angle ADX' = \angle BX'D$, so $2\pi - \angle BAD = 2\angle BX'D = \pi + \angle BCD$. $X$ satisfies this condition, so $X$ lies on $O$, since the arc $BX'D$ is defined by the angle $BX'D$. $\angle BQA = \angle BDA = \frac{\pi}{2} - \frac{\angle BAD}{2} = \frac{\pi}{2} - \angle PAQ$, so $BQ \perp PA$. = Also, $BX \perp PQ$, so $B$, $X$, $Q$ are collinear. Thus, $QX$ is an altitude of $APQ$, and similarly, $PX$ is and altitude of $APQ$. Thus, $X$ is the orthocenter, and we're done.

Let $K$ be a point on $MN$ such that $MB=MK,$ from $MN=BM+DN$ it also follows $ND=NK.$ Say $\angle MBK=\alpha,$ $\angle NDK=\beta,$ so $\angle BCD=180^\circ-2\alpha-2\beta$ $\rightarrow$ $\angle BAD=2\alpha+2\beta$ $\rightarrow$ $\angle ABD=\angle ADB=90^\circ-\alpha-\beta.$ Now, let's show that $K$ is on the circle with center $A$ and radius $AB.$ If $K$ is inside the circle, then $\angle BKD>180^\circ-\alpha-\beta$, and if $K$ is outside the circle, then $\angle BKD<180^\circ-\alpha-\beta.$ But $\angle BKD=180^\circ-\alpha-\beta,$ so, $K$ is on the circle. It means $AB=AK=AD$ $\rightarrow$ $ABMK$ and $ADNK$ are deltoids. So, $AM\perp BK$ and $AN\perp DK.$ Since $BP=KP$ and $\angle BPA=90^\circ-\alpha-\beta,$ we get $\angle MKP=\beta$ $\rightarrow$ $P,K,D$ are collinear. Similarly, we can prove that $Q,K,B$ are also collinear. So, $K,$ which is on $MN,$ is the orthocenter of $\triangle APQ.$

Very nice problem! Solution Let $K$ be a point on the ray $CD$ produced beyond $D$ such that $DK=BM$. It is clear that $\triangle ABM$ and $\triangle ADK$ are congruent. Let the line passing through $B$ perpendicular to $AM$ meet the circumcircle of triangle $CBD$ again at point $T$. We see that $\angle TAM=90^{\circ}-\angle BTA=90^{\circ}-\angle \frac{C}{2}$ and so, $AT$ bisects $\angle MAK$ since $\angle MAK=180^{\circ}-\angle C$. Thus, if $N'=AT \cap CD$ then $MN'=N'K$ since $AT$ is the perpendicular bisector of $MK$. we conclude that $MN'=BM+DN'$ and $N'=N$ follows. Thus, $BP \perp AQ$. Similarly, $DQ \perp AP$. Therefore, $BP \cap QD$ is the orthocenter of triangle $APQ$. By Pascal's Theorem on the hexagon $(CBPAQD)$ we conclude that the orthocenter of $APQ$ lies on $MN$. The result follows.

My solution is different from others. Let the feet of altitudes from $A$ to $CB, CD$ be $X, Y$ respectively. Then $CX=CY=\frac{CM+MN+NC}{2}$ implies $A$ is $C$-excenter of $\triangle CMN$. So $AN, AM$ are internal angle bisector of $\angle MND, \angle NMB$ respectively $(*)$. Let reflections of $C$ across $AP, AQ$ be $C_1, C_2$ respectively. Then by $(*)$, $C_1, C_2\in MN$. It implies that line $MN$ is Steiner line of $C$ respect $\triangle APQ$. So orthocenter of $\triangle APQ$ lies on the line $MN$.

Let the feet of the perpendiculars from $A$ to $\overline{CB}$ and $\overline{CD}$ be $H_{d}$ and $H_{b}$, respectively. Point $A$ lies on the angle bisector of $\angle BCD$, so $AH_{b} = AH_{d}$. Hence $\triangle ABH_{d} \cong \triangle ADH_{b}$, and we get that \[NH_{b} + MH_{b} = DN + MB = NM\]so $A$ is the $C$-excenter for $\triangle CMN$. Therefore $\angle AND = \angle ANM$ and $\angle AMB = \angle AMN$. Now construct a point $H$ on $MN$ such that $NH = ND$, which implies that $AH = AD = AB$ as well. The reflections of $H$ across $\overline{AQ}$ and $\overline{AP}$ (points $D$ and $B$) lie on the circumcircle of $\triangle APQ$, so $H$ is the orthocenter of $\triangle APQ$, which concludes the proof.

Let $K$ be on $MN$ such that $BM=MK$, $DN=NK$. Lemma: $K$ lies on the circle passing through $B$ and $D$ centered at $A$. Proof: It's well known that the incenter $I$ of $BDC$ also lies on this circle. It follows by a simple angle chase that $BIKD$ is concyclic: $BKD=180^{\circ}-\angle DNK-\angle BMK=180^{\circ}-\frac12(\angle CNM+\angle CMN)=90^{\circ}+\frac12(\angle C)=\angle BID$. Then the perpendicular bisectors of $BK,DK$ pass through $A$. So $B$ and $D$ are the reflections of $K$ about $AP$ and $AQ$. Since the orthocenter is the unique interior point whose reflections about the sides lie on the circumcircle, we're done. (Uniqueness is not hard to show, simply observe that the reflections of $(APQ)$ about $AP$ and $AQ$ intersect at two points, one of them being $A$.)