Let $J$ be the midpoint of $AC$ and $DJ$ intersects the perpendicular of $BH$ from $H$ at $K'$,we obtain $HK'\parallel AC$ and hence $\frac{HK'}{DK'}=\frac{AJ}{DJ}=1$,so $K'\equiv K$,as desired. $Q.E.D.$

Let $E$ be the second intersection of circle $K$ with $BC$. $K$ is midpoint of $HE\parallel AC$, hence done. Best regards, sunken rock

http://olympiads.hbcse.tifr.res.in/uploads/crmo-2013-solutions-2 Solution. Note that ∠KHB = 90◦ . Therefore ∠KDA = ∠KHD = 90◦ − ∠BHD =∠HBD = ∠HAC. On the other hand, if M is the midpoint of AC then it is the circumcenter of triangle ADC and therefore ∠MDA = ∠MAD. This proves that D, K, M are collinear and hence DK bisects AC.

Let $H'$ be the $H-$ antipode in $(K)$, easy to see, $H' \in BC$, and $\angle BH'H=90^{\circ}-\angle HBC=C$ $\implies$ $HH'||AC$ and since, $HK=H'K$ $\implies$ $DK$ bisects $AC$

Let $M$ be midpoint of $AC$ We have: $\widehat{HDK}$ = $\widehat{KHD}$ = $\widehat{HBD}$ = $\widehat{HAC}$ = $\widehat{HDM}$ Then: $D$, $K$, $M$ are collinear