In triangle $ABC, BC>AB$. $BL$ are bisector, $BL=AB$. Let $M \in BL$ such that $\angle AML=\angle BCA$. Prove that $AM=LC$. 
In an acute-angled triangle $ABC, BC>AB$. $BL$ are bisector, $BL=AB$. Let $M \in BL$ such that $\measuredangle AML=\measuredangle BCA$. Prove that $AM=LC$.

My work so far:

1)I draw a picture. 
2) If $K \in BC$, such that $BK=BL$ then $\triangle BAL=\triangle BLK$. But $\triangle AML\not=\triangle CLK$
 A: Note that $\measuredangle AML=\measuredangle BCA$, and $\measuredangle MLA= \measuredangle BAC$, so we have that $$\triangle AML \sim \triangle BCA \tag{1}$$
Now, use the angle bisector theorem to find that $$\overline{AB}: \overline{AL}=\overline{BC}:\overline{CL} \tag{2}$$
So we have $$\overline {AM} \times \frac{\overline{AB}}{\overline{AL}}= \overline{BC}=\overline{CL} \times \frac{\overline{BC}}{\overline{CL}}=\overline{CL} \times \frac{\overline{AB}}{\overline{AL}}$$
From $\text{(1)}$ and $\text{(2)}$. 
Our proof is done. 
A: Or, alternatively, take the point $B'$ on line $BC$ such that $B'L=BL$. (For instance draw the circle centered at $L$ and passing through point $B$. Then $B'$ is the other point of intersection of the circle with line $BC$.) 

Then triangle $LBB'$ is isosceles and therefore $\measuredangle \, MBA = \measuredangle \, LBC = \measuredangle \, CB'L$. Furthermore, $\measuredangle \, LCB' = \measuredangle \, AMB = 180^{\circ} - \measuredangle\, BCA =  180^{\circ} - \measuredangle\, AML$. Combined with the fact that $AB = BL = B'L$ these conclusions yield that triangles $ABM$ and $LB'C$ are congruent and therefore $LC = AM$.  
