Let $KH$ and $DC$ intersect at point $T$. Let $X$ be the foot of the perpendicular $D$ onto $AB$. $DX^2=DH^2=DT.DC$ and therefore $DX$ is tangent to circle $XTC$. The center of circle $XTC$ is the intersection of perpendicular bisector of $CT$ and line passing through $X$ and perpendicular to $DX$, which is clearly $M$.

Trig ! Let $AM=MK=x$ , $BK=y$ and $BH=z$ . Also , as usual let $BC=a , CA =b , AB = c $ Note that $2x+y=c$ . We now compute $z$ . Note that,we have $z=BD\sin{\frac B2}$ . By Law of Sines in $\Delta BDC$ ,we must have $$ \frac{BD}{\sin\left(\frac{\pi}{2}-C\right)}=\frac{a}{\sin\left(C+\frac{B}{2}\right)} \iff BD = \frac{a\cos C}{\sin\left(C+\frac {B}{2}\right)}$$ So we have $z=\frac{a\sin{\frac{B}{2}} \cos C}{\sin\left(C+\frac{B}{2}\right)}$ Also , note that $$\frac{y}{z} = \frac{c}{a} \iff y= \frac{c\sin{\frac{B}{2}}\cos C}{\sin\left(C+\frac B2\right)}$$. By law of cosines in $\Delta AMC$ , we need $$(x+y+z)^2 = b^2+x^2-2bx\cos A$$$$\iff (2x+y+z)(y+z)=b[b-2x\cos A]$$$$\iff (c+z)(y+z)=b[b-(c-y)\cos A]$$$$\iff (c+z)(y+z)=b [b- c \cos A + y \cos A]$$$$\iff (c+z)(y+z)=b[y\cos A+a\cos C]$$ This is just 15 minute calculation and is left to the reader

i have learnt. solved with Om245 and rjp08