Geometric proof: Legs intersect on CM (median of triangle) $M$ is the midpoint of $AB$ in the triangle $\triangle ABC$. The angle $\angle ACM$ is copied and drawn on the leg $AB$ in $A$. The angle $\angle MCB$ is copied and drawn on the leg $BA$ in $B$. The direction of rotation is chosen so that the free legs are on the same side of AB like the point $C$. 

Prove that the free legs always intersect on the line $\overline {CM}$.

I drew the whole figure for an arbitrary triangle. For a better understanding: the original triangle was $\triangle ABC$ and then I created the triangle $\triangle ABC_1$ with the angles Alpha and Beta.
I tried some different approaches (like Ceva's theorem or Menelaus' theorem, even a trigonometric approach or with vectors) but I am really stuck on this task. It occurs to me that I have too little information for this task, but maybe one of you has a better approach.
Any kind of help or advice will be really appreciated.

 A: It's easy. Let these two lines intersect at $X$. Let $CM$ intersect the circumcircle of triangle $ABC$ at point $D$. Then $\angle ACM = \angle ACD = \angle ABD = \angle BAX$, so $AX \parallel BD$. Analogously we show that $BX \parallel AD$. Thus $ADBX$ is a parallelogram which means that $D,X,M$ are colinear. On the other hand $C,M,D$ are collinear (by definition of $D$). Thus $C,D,M,X$ are collinear and we are done.

A: Let $\Gamma_B$ be the circumcircle of $AMC$, $\Gamma_A$ be the circumcircle of $CMB$ and $\Gamma_M$ be the circle having $AB$ as a diameter. Let $\Gamma_B\cap \Gamma_M=\{A,A'\}$, $\Gamma_A\cap\Gamma_M=\{B,B'\}$.

Since $MA'=MA$ we have $\widehat{MAA'}=\widehat{ACM}$ and since $MB=MB'$ we have $\widehat{B'BM}=\widehat{MCB}$. 
In this configuration, $CM,AA'$ and $BB'$ concur in the radical centre of $\Gamma_A,\Gamma_B,\Gamma_M$, proving your claim. Here it is the other configuration, in which $\widehat{C}$ is an obtuse angle:

A: 
Let $P$ and $Q$ be the (possibly-distinct, possibly-coincident) points where the respective segments from $A$ and $B$ meet $\overline{CM}$. Certainly,
$\angle APM\cong\angle CAM$, and $\angle BQM\cong\angle CBM$; thus, $\triangle APM \sim \triangle CAM$, and $\triangle BQM \sim \triangle CBM$, and we have
$$\begin{align}
\frac{|\overline{PM}|}{|\overline{AM}|} = \frac{|\overline{AM}|}{|\overline{CM}|} &\quad\to\quad |\overline{PM}| = \frac{|\overline{AM}|^2}{|\overline{CM}|} \\[6pt]
\frac{|\overline{QM}|}{|\overline{BM}|} =\frac{|\overline{BM}|}{|\overline{CM}|} &\quad\to\quad |\overline{QM}| = \frac{|\overline{BM}|^2}{|\overline{CM}|}
\end{align}$$
Since $\overline{AM}\cong\overline{BM}$, it follows that $\overline{PM}\cong\overline{QM}$. Points $P$ and $Q$ coincide. $\square$
(Note that the roles of point $C$ and point $P$/$Q$ are interchangeable here, since the construction is reversible.)
