Concyclic points on a circle This is a question I met in some notes. I need some help here.

Let $ABC$ be an acute triangle and let $AA_1, BB_1, CC_1$ be its altitudes. Segments $AA_1$ and $B_1C_1$ meet at $K$. The perpendicular bisector of segment $A_1K$ intersects sides $AB$ and $AC$ at $L$ and $M$ respectively. Prove that points $A, A_1, L$ and $M$ lie on a circle.

Can anyone help?
 A: For reasons of (parallel lines, angles in the same segment, and properties of orthic triangle), all the blue marked angles are equal. In particular $\phi = \phi’$.

Let LM meet $BB_1$ at N’, and $CC_1$ at N. Since LN’NM is the perpendicular bisector of $KA_1$, a circle can be drawn through K, L, $A_1$ with its diameter being part of LM. $\angle LC_1N = 90^{\circ}$ confirms that LN is its diameter.
Added: A circle passing through $K, L, A_1$ can definitely be formed. Note that  LM is the perpendicular bisector of $KA_1$. This means the center of that circle is somewhere on LM. This also means the diameter of that circle lies on LM. By converse of "angles in semi-circle", if $\angle LC_1N = 90^{\circ}$, then LN is the diameter of that circle. 
This means N, K, $C_1$, L, $A_1$ are con-cyclic. Therefore, $\phi’’ = \phi = \phi’$.
Required result follows.
A: 
In a triangle $ABC$ with orthocenter $H$, $h_a = a/(\cot \beta + \cot \gamma)$ and $AH = a \cot \alpha$.
From $h_c/AC_1 = h_b/AB_1 = \tan \alpha$ and $c h_c = b h_b$, we have $\triangle AB_1C_1 \sim \triangle ABC$. Then, from $AC_1/AH = \cos \alpha_1$ and from $\triangle AKC_1$,
$$\frac {AK} {\sin \gamma} =
\frac {AC_1} {\sin(\gamma + \alpha_1)} =
\frac {AH \sin \beta} {\sin \gamma \sin \beta + \cos \gamma \cos \beta}, \\
AK = \frac {a \cot \alpha} {1 + \cot \beta \cot \gamma}.$$
From $\triangle ABC \sim \triangle ALM$,
$$\frac {LM} a = \frac {AN} {h_a} = \frac {(h_a + AK)/2} {h_a} =
\frac 1 2 \left( 1 -
 \frac {\cot(\beta +\gamma)(\cot \beta + \cot \gamma)}
  {1 + \cot \beta \cot \gamma} \right) =
\frac 1 {1 + \cot \beta \cot \gamma}, \\
AK = LM \cot \alpha.$$
Therefore, $K$ is the orthocenter of $\triangle ALM$. Since $AC'KB'$ is cyclic and $\angle C'KB' = \angle LA_1M$, $ALA_1M$ is also cyclic.
