It should be "median BM" , not "AM"

Redefine $D$ as the reflection of $C$ over $BM$; note that $AD\perp CD\perp BM$. Now $\angle BDM=\angle BCM=\angle CHM$, so $D$ lies on $(BHM)$.

Let $\Delta ABC$ and $\Delta DEF$ be two \(\triangle\) such that $\angle A = \angle D ; BC = EF$ $; A -altitude \Delta ABC = D - altitude \Delta DEF$ $\Rightarrow \Delta ABC \cong \Delta DEF$ Returning to our original figure. Same is the case with \(\Delta BCM\) and \(\Delta BDM\) in our figure.

[asy][asy]size(6cm); pair A=(3,8),B=(0,0),C=(10,0),H,M,D,E; H=foot(A,B,C);M=(A+C)/2;E=A+B-M;D=reflect(B,M)*C; D(circumcircle(B,H,M),green); D(B--MP("D",D,NE)--M--B--MP("E",E,W)--A--H--M--E--H,magenta);D(A--D,magenta); D(MP("A",A,N)--MP("B",B,S)--MP("H",H,S)--MP("C",C,S)--MP("M",M,NE)--A); dot(A^^B^^C^^H^^M^^E^^D); [/asy][/asy] Complete the parallelogram $AMBE$; clearly $EM$ is the $A-$midline in $\triangle ABC$. So $H$ is the reflection of $A$ over $EM$, so $\angle EBM=\angle EAM=\angle EHM$, which means $E\in\odot(BHM)$. Now clearly $EMCB$ is a paralleogram too ($EM||BC$ and $EB||MC$), implying $ME=BC$. But $ED||BM$, so $EDMB$ is an isosceles trapezium, which immediately gives $BD=ME=BC$. $\blacksquare$