Given an acute triangle ABC with altitudes AH, BK. Let M be the midpoint of AB Given an acute triangle ABC with altitudes AH, BK. Let M be the midpoint of AB. The line through CM intersect HK at D. Draw AL perpendicular to BD at L. Prove that the circle containing C, K and L is tangent to the line going through BC


This proof does not assume that $ABC$ is acute.  Suppose $a:=BC$, $b:=CA$, and $c:=AB$.  Likewise, $\alpha:=\angle BAC$, $\beta:=\angle ABC$, and $\gamma:=\angle BCA$, so $\alpha+\beta+\gamma=\pi$.  Let $x:=\angle CKL$ and $z:=\angle AMC$.  Note that $A$, $K$, $L$, $H$, and $B$ lie on the circle centered at $M$ with radius $\frac{c}{2}$.
Note that $$\angle KBL=\angle ABL-\angle ABK=\angle CKL-\left(\frac{\pi}{2}-\alpha\right)=x+\alpha-\frac{\pi}{2}\,,$$
$$\angle LBC=\angle ABC-\angle ABL=\beta-\angle CKL=\beta-x\,,$$
$$\angle BCM=\angle AMC-\angle ABC=z-\beta\,,$$
$$\angle MCK=\pi-\angle CAM-\angle AMC=\pi-\alpha-z\,,$$
$$\angle CKH=\angle ABC=\beta\,,$$
and 
$$\angle HKB=\angle LAB=\frac{\pi}{2}-\beta\,.$$
 The lines $BL$, $CM$, and $KH$ concur.    By the trigonometric version of Ceva's Theorem on the triangle $BCK$, we have
$$1=\frac{\sin(\angle KBL)}{\sin(\angle LBC)}\cdot\frac{\sin(\angle BCM)}{\sin(\angle MCA)}\cdot\frac{\sin(\angle CKH)}{\sin(\angle HKB)}=-\frac{\cos(x+\alpha)}{\sin(\beta-x)}\cdot\frac{\sin(z-\beta)}{\sin(\alpha+z)}\cdot\frac{\sin(\beta)}{\cos(\beta)}\,.$$
That is,
$$
\begin{align}
\frac{\cos(\beta)\tan(x)-\sin(\beta)}{\cos(\alpha)-\sin(\alpha)\tan(x)}&=-\frac{\sin(\beta-x)}{\cos(x+\alpha)}=\frac{\sin(z-\beta)}{\sin(\alpha+z)}\cdot\frac{\sin(\beta)}{\cos(\beta)}
\\
&=\frac{\sin(\beta)}{\cos(\beta)}\cdot\frac{\tan(z)\cos(\beta)-\sin(\beta)}{\sin(\alpha)+\cos(\alpha)\tan(z)}\,.
\end{align}$$
Let $F$ be the feet of the perpendicular from $C$ to $AB$.  We have $CF=b\sin(\alpha)$ and $MF=\frac{c}{2}-b\cos(\alpha)$ ($MF$ is taken to be a signed length, so $MF$ is negative if $\alpha<\beta)$.  That is, $$\tan(z)=\frac{b\sin(\alpha)}{\frac{c}{2}-b\cos(\alpha)}=\frac{2\sin(\alpha)\sin(\beta)}{\sin(\gamma)-2\cos(\alpha)\sin(\beta)}\,,$$
since $\frac{b}{\sin(\beta)}=\frac{c}{\sin(\gamma)}$ due to the Law of Sines on the triangle $ABC$.  Now, $\gamma=\pi-\alpha-\beta$, so we get $\sin(\gamma)=\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$.  Therefore,
$$\tan(z)=\frac{2\sin(\alpha)\sin(\beta)}{\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)}\,.$$
Consequently,
$$
\begin{align}
\frac{\cos(\beta)\tan(x)-\sin(\beta)}{\cos(\alpha)-\sin(\alpha)\tan(x)}&=
\frac{\sin(\beta)}{\cos(\beta)}\cdot\frac{\tan(z)\cos(\beta)-\sin(\beta)}{\sin(\alpha)+\cos(\alpha)\tan(z)}
\\
&=\frac{\sin(\beta)}{\cos(\beta)}\cdot\frac{2\sin(\alpha)\sin(\beta)\cos(\beta)-\sin(\beta)\big(\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)\big)}{\sin(\alpha)\big(\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)\big)+2\sin(\alpha)\cos(\alpha)\sin(\beta)}
\\
&=\frac{\sin^2(\beta)}{\sin(\alpha)\cos(\beta)}\cdot\frac{\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)}{\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)}=\frac{\sin^2(\beta)}{\sin(\alpha)\cos(\beta)}\,.
\end{align}$$
(Technically, we have to worry about the case $\alpha=\beta$, but we can argue by continuity that in the limit $\alpha=\beta$, the above equality still holds.)
Ergo,
$$\sin(\alpha)\cos^2(\beta)\tan(x)-\sin(\alpha)\sin(\beta)\cos(\beta)=\cos(\alpha)\sin(\beta)^2-\sin(\alpha)\sin^2(\beta)\tan(x)\,,$$
leading to
$$\sin(\alpha)\tan(x)=\sin(\beta)\big(\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\big)=\sin(\beta)\sin(\alpha+\beta)=\sin(\beta)\sin(\gamma)\,.$$
That is,
$$\tan(x)=\frac{\sin(\beta)\sin(\gamma)}{\sin(\alpha)}=\frac{c\sin(\beta)}{a}=\frac{AH}{BC}\,,$$
where we have once again used the Law of Sines $\frac{a}{\sin(\alpha)}=\frac{c}{\sin(\gamma)}$.
Now, as $\angle ABL=\angle CKL=x$, we have
$$\frac{AL}{BL}=\tan(x)=\frac{AH}{BC}\,.$$
The triangles $AHL$ and $BCL$ have $\angle LAH=\angle LBH=\angle LBC$ and $\frac{AL}{BL}=\frac{AH}{BC}$.  Therefore, $AHL$ and $BHL$ are similar triangles, whence $\angle BCL=\angle LHC$.  However, $\angle{LHC}=\angle LKC=x$.  Thus, $\angle BCL=\angle LKC=x$.  This means $BC$ is tangent to the circumscribed circle of the triangle $CKL$.
P.S.  As a result of this problem, we can also show that the circumscribed circle of $BCL$ is tangent to $AB$ and that, if $N$ is the midpoint of $CH$, then $MN$ is a perpendicular bisector of $HL$, and that the circle with diameter $CH$ passes through $L$.
