Let $T$ be the reflection of $M$ over $A$. We have that $AK$ is parallel to $TN$ cause it's the midline of $\triangle MNT$. Also, $DN$ is parallel to $AK$, because $AKND$ is a parallelogram. Hence $T-D-N$ lie on the same line, moreover $TD = DN = AK$. Let $C'$ be a point on $BC$ such that $DC \| TC'$. $D$ is the midpoint of $TN$, hence $C$ is the midpoint of $C'N$. Now angle chasing : $\angle DAB = \angle TC'B = \pi - \angle AMN$, hence $TMNC'$ is cyclic. Also $H$ is the anti-center of this quadrilateral($CH$ is perpendicular to $AB$ and $AH$ is perpendicular to $BC$). This means that $KH \perp TC'$. q.e.d [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.337966298044838, xmax = 3.1177401710314734, ymin = -3.4201936533156188, ymax = 5.782332492033476; /* image dimensions */ /* draw figures */ draw((-4.5486987840077795,2.212214504768269)--(-5.31044047679765,1.1310154770491578), linewidth(0.5)); draw((-4.8600200773151885,1.6226149521152684)--(-4.999119183490241,1.7206150297021585), linewidth(0.5)); draw((-0.5406481605835146,-0.8504262355928098)--(-4.5486987840077795,2.212214504768269), linewidth(0.5)); draw((-5.31044047679765,1.1310154770491578)--(-4.071629875554594,-2.2870882481740122), linewidth(0.5)); draw((-4.071629875554594,-2.2870882481740122)--(-0.5406481605835146,-0.8504262355928098), linewidth(0.5)); draw((-4.5486987840077795,2.212214504768269)--(-4.071629875554594,-2.2870882481740122), linewidth(0.5)); draw((-4.5486987840077795,2.212214504768269)--(-2.9103052981225925,-1.8145763409048634), linewidth(0.5)); draw((-4.071629875554594,-2.2870882481740122)--(-2.076870544034293,0.32343549916525544), linewidth(0.5)); draw((-0.5406481605835146,-0.8504262355928098)--(-4.183011542184318,-1.2366323697574473), linewidth(0.5)); draw((-4.071629875554594,-2.2870882481740122)--(-2.61967364099556,-0.9250593853734639), linewidth(0.5)); draw((-2.61967364099556,-0.9250593853734639)--(-5.31044047679765,1.1310154770491578), linewidth(0.5)); draw((-3.3814153337854305,-2.0062584130925747)--(-7.239465619809869,4.268289367190891), linewidth(0.5)); draw((-7.239465619809869,4.268289367190891)--(-4.5486987840077795,2.212214504768269), linewidth(0.5)); draw(circle((-5.431962926581172,1.0562945985790089), 3.685644635558221), linewidth(0.5)); draw((-4.761844417323746,-2.5679180832554462)--(-4.071629875554594,-2.2870882481740122), linewidth(0.5)); draw((-5.039419208333393,-1.8020387581785704)--(-2.61967364099556,-0.9250593853734639), linewidth(0.5)); draw((-7.239465619809869,4.268289367190891)--(-4.761844417323746,-2.5679180832554462), linewidth(0.5)); draw((-3.3814153337854305,-2.0062584130925747)--(-2.61967364099556,-0.9250593853734639), linewidth(0.5)); draw((-3.070094040478021,-1.416658860439575)--(-2.930994934302968,-1.5146589380264635), linewidth(0.5)); draw((-2.61967364099556,-0.9250593853734639)--(-1.8579319482056895,0.15613964234564698), linewidth(0.5)); draw((-2.308352347688152,-0.3354598327204639)--(-2.1692532415130987,-0.4334599103073525), linewidth(0.5)); draw((-4.5486987840077795,2.212214504768269)--(-2.61967364099556,-0.9250593853734639), linewidth(0.5)); /* dots and labels */ dot((-4.5486987840077795,2.212214504768269),dotstyle); label("$A$", (-4.496676685751994,2.350882064954153), NE * labelscalefactor); dot((-5.31044047679765,1.1310154770491578),dotstyle); label("$D$", (-5.802681368860207,1.0321841735559005), NE * labelscalefactor); dot((-0.5406481605835146,-0.8504262355928098),dotstyle); label("B", (-0.48386482267988734,-0.7119001344224334), NE * labelscalefactor); dot((-4.071629875554594,-2.2870882481740122),dotstyle); label("$C$", (-4.028751627513904,-2.6236003673747916), NE * labelscalefactor); dot((-3.3814153337854305,-2.0062584130925747),dotstyle); label("$N$", (-3.3764924554244446,-2.2574707813300625), NE * labelscalefactor); dot((-1.8579319482056895,0.15613964234564698),dotstyle); label("M", (-1.8025627140781415,0.29484771815042604), NE * labelscalefactor); dot((-2.61967364099556,-0.9250593853734639),dotstyle); label("K", (-2.7500537361156496,-0.7827978705191136), NE * labelscalefactor); dot((-3.188365732981202,-1.131168341984753),dotstyle); label("$H$", (-3.3354401526957316,-0.9954910788091543), NE * labelscalefactor); dot((-4.183011542184318,-1.2366323697574473),dotstyle); dot((-2.9103052981225925,-1.8145763409048634),dotstyle); dot((-2.076870544034293,0.32343549916525544),dotstyle); dot((-7.239465619809869,4.268289367190891),dotstyle); label("$T$", (-7.17661111020651,4.40691641175788), NE * labelscalefactor); dot((-5.039419208333393,-1.8020387581785704),dotstyle); dot((-4.761844417323746,-2.5679180832554462),dotstyle); label("$C'$", (-4.851165366235396,-2.9671913117615127), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] PS. I didn't solve the problem during the TST

https://artofproblemsolving.com/community/c6h1633431p10262843

khina wrote: https://artofproblemsolving.com/community/c6h1633431p10262843 LMFAO. Moldova TST moment

we 'll use directed angles ; consider $C'$, the symmetric of $C$ in $DK ,L=BN\cap KD; H,C,C'$ are collinear; it s clear that $DCKN$ is cyclic so $\angle DCK=\angle DNK\implies \angle DC'K=DAK$ i.e $AKC'D$ is cyclic ; $\angle AKC'=\angle AKD+\angle DKC'=\angle AKD+\angle CKD=\angle NDK+\angle CND=\angle DLC=\angle LBM$ but $HA\perp BC, HC\perp BM$ then $\angle AKC'=\angle AHC' $ hence $ AHDKC'$ is cyclic more we have $HC \perp KD$ and $\angle DHK=\angle DCK$ hence $ C$ is the orthocenter of $ HDK$ therefore the result follows . My regards. RH HAS