TITLE: Prove that line through intersection of line through foots of heights and opposite line and orthocenter is perpendicular to median.
QUESTION [15 upvotes]: $BB'$ and $CC'$ are heights of a given $\triangle ABC$ ($AB\ne AC$). $M$ is mid-point of $BC$ and $H$ is orthocenter of $\triangle ABC$ and $D$ is intersection of lines $B'C'$ and $BC$. Prove $DH \perp AM$.


My idea is to prove that $AC'EHB'$ (or $C'BME$ or $B'EMC$) is cyclic. I tried lots of things, but I think the main thing I lacked in all my deductions if usage of $M$ being a mid-point.
Other idea is to prove that $H$ is orthocenter of $\triangle AMD$, since we already have $AH\perp MD$, it's enough to prove $MH\perp AD$. If $MH\cap AD=\{G\}$, then we should prove that $G$ lies on circle around $AC'(E)HB'$. So another idea would be to prove that some of triangles "composed" of $A,C',E,H,B',G$ with at least once vertices being $E$ or $G$ has circumradius $AH/2$. Some trigonometry?
Another idea would be to mark $E'$ as foot of height from $D$ to $AM$ and then use Menelaus theorem to prove that $A-H-E'$, that is, $E\equiv E'$.
EDIT. The only place where I think I can use the fact that $M$ is midpoint of $BC$ is that $BM=C'M=B'M=CM$, so there are some equal angles. It should help us prove $E$ lies on circumcircle of $\triangle C'BM$, that is, $C'BME$ is cyclic, because we have $\angle MC'B=MBC'$. It's also interesting to notice that once we prove $E$ lies on mentioned circle, we'd conclude that $DE$ is bisector of $\angle C'EB$, which might be a clue but I still can't figure anything out.
EDIT 2. I also thought about using the fact that $\triangle CMA$ and $\triangle BMA$ have same areas. Then we have $AC\sin\alpha_1=BC\sin\alpha_2$ where $\alpha_1=\angle CAM=\angle CAE$ and $\alpha_2=\angle BAM=\angle BAE$, so we might be able to use that somehow for the idea about circumradius being $AH/2$, because we can use sine laws on triangles $\triangle AC'E$ and $AEB'$ (we need to prove $\angle C'EA=\angle B=\beta$ and/or $\angle B'EA=\angle C=\gamma$), from which we might reduce this to proof about similarity of some triangles. Seems like a nice idea that could work but I can't seem to get much from it for now.
Another thing, about the idea of proving that $H$ is orthocenter of $\triangle AMD$. I'm not sure is it better to mark $MH\cap AD=\{G\}$ and prove $MH\perp AD$, or other way around? Or maybe to mark intersection of $AD$ and circumcirle of $AC'HB'$ and $\triangle ABC$ (maybe this is important?) as $G$ and then prove $H\in GM$?
Just adding ideas, since I can't seem to get anywhere, almost like we lack just a little bit something and all of the ideas might actually work very similarly.
Also, since people wondered if it's Olympiad problem or homework problem: it's from Serbian mathematical magazine for high schoolers, it's in the section where one of the hardest problems are. But I still find it harder than the usual ones from geometry.
Please no analytic geometry or complex numbers, and try avoiding vectors unless used only for less significant lemmas.

REPLY [3 votes]: this problem is really orthogonal circles, let us from pure geometry view to solve it. 
first to find two orthogonal circles from triangle ABC. see picture below:

look BC'B'C: $ {CC'} \perp {AB} $,$BB' \perp AC$,so BC'B'C is cyclic and M is the center.
look at AC'HB',they are also cyclic. the center is the N of midpoint of AH.
BTW,A'C'NBM is cyclic which is the 9 points circlehttp://en.wikipedia.org/wiki/Nine_points_circle and $C'N \perp C'M $ as NM is the diameter.
so circle NA and circle MC are orthogonal circles.
2nd: we will proof if DM is the diameter to make a circle, then this circle and circle NA are also orthogonal circles. I use another picture to proof it.

suppose circle A and B are orthogonal circles, C and D are the two common points, then $AD \perp BD $,$AC \perp BC$. line AB,CD, cross at E. then $AB \perp CD$. let $AD=p,BD=q$, then $AB= \sqrt{p^2+q^2}=d $
take F on CD,$ FE=x, AE=m= \dfrac{p^2}{d}, BE=n=\dfrac{q^2}{d}$, line AF, G is midpoint of AF, $GF=r=\dfrac{AF}{2}=\dfrac{\sqrt{x^2+m^2}}{2}$,make $ GH \perp AB,$ foot H, then $GH= \dfrac{FE}{2}=\dfrac{x}{2},HE=\dfrac{m}{2},GB^2=GH^2+HB^2=(\dfrac{x}{2})^2+(\dfrac{m}{2}+n)^2=\dfrac{x^2}{4}+\dfrac{m^2}{4}+mn+n^2)=r^2+n(m+n)=r^2+nd=r^2+q^2$.
so circle GF and circle B are orthogonal circles. look at diameter JK of circle A, please note A is midpoint of JK.BTW, circle G(red) must through E.
3rd: orthogonal circles have some interesting fact which already proof, see another picture:

circle P and Q are orthogonal, then take A in circle P, make AB is diameter,E and F are common points of circle P and Q, AE meet circle Q at C, then $ CQ \perp AB$,and CQ cross circle Q at D, BC must through E, D must the orthocenter of ABC. 
I left detail proof as exercise , you can find details from "Notes On the Orthcentric Tetrahedron" paper on web ,which is recommended in SE. 
last step: see final picture:

K is midpoint of DM, make circle KM, then circle KM and circle NA are orthogonal circles. look at triangle ADM, （you may need to proof A must be the point which equal to 3rd step point C,it is not difficult when you finish 3rd step.)$AN \perp DM $, cross circle NA　at H, so H must be the orthocenter　of ADM. that is done.
I add detail proof for the last step:
circle KD have common points E and F. line KE,NE, since two circles are orthogonal circles which ii proof in step2, so $KE \perp NE$,line ME ,suppose cross at circle NA at A2, line A2N, $\angle NA2E=\angle NEA2,\angle KEM=\angle KME, \angle KEM+\angle NEA2=\dfrac{\pi}{2},$ so $\angle KME+\angle NE2A=\dfrac{\pi}{2} $,that is $A2N \perp DM$, since both AN and A2N have common point N, A2 must be A. with same reason, DF extend circle NA, it must cross at A. if we connect DE,HE,since DM is diameter, so $ DE\perp AM$, since AH is diameter, so $HE \perp AM$, both have common point E, that means H must on DE. and H must orthcenter of triangle ADM. that is finished.