HINT : It's easy by Trigo. Just take $AB=1,\angle {AMB}=x,\angle {CMD}=y$ Now take projection of $H_x,H_y,H_z$ on $BC$ as $H'_x,H'_y,H'_z$ .Also $H_xH'_x=a,H_yH'_y=c,H_zH'_z=b,H'_xH'_z=p,H'_yH'_z=q$ Now it's enough to show $pa(b-c)=(b-a)b(p+q)$ .By simple Using of Sine rule from those three triangles we'll get that result.

I also know the metric solution but i am looking for the Synthetic as $H_z$ seems to have some weird properties.

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paul1703 wrote: I also know the metric solution but i am looking for the Synthetic as $H_z$ seems to have some weird properties. http://geometry.ru/olimp/sharygin/2012/fineng2.pdf

Dear Mathlinkers, any ideas for a synthetic proof? Sincerely Jean-Louis

This problem true also for rhombus $ABCD$.

Dangit, this is a problem from Sharygin olympiad(final round). I have solved this before, but I can't remember the previous solution. Denote $J, K$ as the intersection of $AM, BH_x$ and $DM, CH_y$. By using Menelaus like a maniac we have $AJ, BK, H_xH_y$ are concurrent. Furthermore, we easily obtain $A, X, K, H_x$ are concyclic points and $D, Y, J, H_y$ are concyclic too. From here, by angle chasing we have $AK=DJ=AD$. But since $AZ, DZ$ are bisector of $\widehat{DAM}$ and $\widehat{MDA}$ hence $AK \perp DZ$ and $DJ \perp AZ$. Hence proved. P/s:It seems that this solution works for buratinogigle's generalization

paul1703 wrote: A point $M$ lies on the side $BC$ of square $ABCD$. Let $X$, $Y$ , and $Z$ be the incenters of triangles $ABM$, $CMD$, and $AMD$ respectively. Let $H_x$, $H_y$, and $H_z$ be the orthocenters of triangles $AXB$, $CY D$, and $AZD$. Prove that $H_x$, $H_y$, and $H_z$ are collinear. Let $K,L$ be points on $AM, DM$ such that $AK = DK = AB$. Note that $H_x = BK \cap AC$, $H_y = CL \cap BD,$ $H_z = AL \cap BK$. Note that $\triangle BKD$ and $\triangle ALC$ are in perspective as $M = BC \cap AK \cap DL$. So, by Desargues theorem, $H_x, H_y, H_z$ are collinear. $\blacksquare$