Law of Sines in Triangle On the side BC of the triangle ABC we construct towards the exterior a square BCDE. Denote the intersection between AE and BC by M. Use the law of sines to prove that 
$$\frac{BM}{CM}=\frac{\cos \measuredangle B\cdot \sin \measuredangle C}{\sqrt{2} \cdot \sin \measuredangle B \cdot \sin(\measuredangle C+45°)}$$
If someone could please help me prove this problem. I do not have a similar problem to work off of. I am unclear of where the $\sin(\measuredangle C+45^{\circ})$ comes into play and the $\sqrt{2}$. 
 A: 
Using sine rule at $ACE$ we have:
$$\frac{AE}{\sin(C+45°)}=\frac{AC}{\sin \alpha}$$
Using sine rule at $ABE$ we have:
$$\frac{AE}{\sin(B+90°)}=\frac{AB}{\sin \beta}$$
Dividing both equations we have:
$$\frac{\cos B}{\sin (C+45°)}=\frac{AC}{AB}\cdot \frac{\sin \beta}{\sin \alpha} \quad (1)$$
Using sine rule at $ABC$ we get
$$\frac{AC}{AB}=\frac{\sin B}{\sin C}$$
So, from $(1)$
$$\frac{\cos B\cdot \sin C}{\sin B\cdot \sin (C+45°)}=\frac{\sin \beta}{\sin \alpha}\quad (2)$$
but $\alpha + \beta=45°$ so
$$\frac{\sin \beta}{\sin \alpha}=\frac{\sin \beta}{\sin (45°-\beta)}=\sqrt{2}\cdot\frac{\tan \beta}{1-\tan \beta}\quad (3)$$
Finaly we can use, from the triangle BME, that
$$\tan \beta = \frac{BM}{BE}=\frac{BM}{BM+CM}\to \frac{BM}{CM}=\frac{\tan \beta}{1- \tan \beta} \quad (4)$$
Pluging $(3)$ and $(4)$ at $(2)$ we get what we want
A: Let $\measuredangle B=\beta$, $\measuredangle C=\gamma$, $\measuredangle BAE=\alpha_1$, $\measuredangle CAE=\alpha_2$. By applying the sine rule to triangles $BAM$ and $CAM$ one gets:
$$
{BM\over\sin\alpha_1}={AM\over\sin\beta},\quad
{CM\over\sin\alpha_2}={AM\over\sin\gamma},\quad
\hbox{whence:}\quad
{BM\over CM}={\sin\alpha_1\over\sin\alpha_2}{\sin\gamma\over\sin\beta}.
$$
By applying then the sine rule to triangles $BAE$ and $CAE$ one gets:
$$
{\sin\alpha_1\over BE}={\sin(\beta+90°)\over AE},\quad
{\sin\alpha_2\over CE}={\sin(\gamma+45°)\over AE}.
$$
From that, taking into account that $CE=\sqrt2 BE$, one can readily obtain 
$\displaystyle{\sin\alpha_1\over\sin\alpha_2}$ and thus the desired result.
A: In triangle $ABM$, using the law of sines implies
$$\frac{BM}{AM}=\frac{\sin\measuredangle BAM}{\sin\measuredangle B}$$
similarly, in $ACM$:
$$\frac{CM}{AM}=\frac{\sin\measuredangle CAM}{\sin\measuredangle C}$$
combining these two yields:
$$\frac{BM}{CM}=\frac{\sin\angle BAM}{\sin\angle CAM}\cdot\frac{\sin\angle C}{\sin\angle B}\label{*}\tag{*}$$
Now, in triangle $ABE$ note that $\measuredangle ABE=\measuredangle B+90^{\circ}$. 
Thus $\sin\measuredangle ABE=\cos\measuredangle B$ and the law of sines implies
$$\frac{\sin\measuredangle BAM}{\cos\measuredangle B}=\frac{BE}{AE}$$
and for triangle $ACE$:
$$\frac{\sin\measuredangle CAM}{\sin(\measuredangle C+45)}=\frac{CE}{AE}$$
Note that $CE=\sqrt{2}BE$. Now combining the two equations implies:
$$\frac{BE}{CE}=\frac 1{\sqrt 2}=\frac{\sin\measuredangle BAM}{\sin\measuredangle CAM}\cdot\frac{\sin(\measuredangle C+45)}{\cos\measuredangle B}$$
And finally, you can use $\eqref{*}$ to get the desired result.
