What is the value of the $\measuredangle x$ in the figure below? For reference:
If ABCD is a square calculate "x" being "M" and "N" midpoints of CD and AD respectively.

My progress:
$\triangle ABN \cong \triangle BC (2:1- special~ right~ triangle) \implies \measuredangle ABN = \frac{53^o}{2}=\measuredangle MBC\\
\therefore \measuredangle NBM = 90-53 = 47^o$
is there any property that shows that $x$ and $47^o$ are complementary?

 A: 
WLOG, we assume $ABCD$ is a unit square. Then,
$BN = NP = \frac{\sqrt5}{2}$
Using power of point of $B$,
$BH \cdot BP = AB^2 \implies BH \cdot \sqrt5 = 1$
$BH = \frac{1}{\sqrt5}$
Now, $\frac{BQ}{BR} = \frac{HQ}{NR} = \frac{BH}{BN}$
$\frac{BQ}{3 / 2 \sqrt2} = \frac{HQ}{1/2\sqrt2} = \frac{1/\sqrt5}{\sqrt5/2}$
$BQ = \frac{3}{5 \sqrt2}, HQ = \frac{1}{5 \sqrt2}$
$GQ = BG - BQ = \frac{1}{\sqrt2} - \frac{3}{5 \sqrt2} 
 = \frac{2}{5 \sqrt2}$
In right triangle $\triangle GQH$, perpendicular sides are in ratio $1:2$, so $\angle HGQ \approx 26.5^\circ$
$\therefore x \approx 53^\circ$
A: I favor the less elegant approach of working entirely in Cartesian coordinates.  Assume that the coordinates of points A and C are $(0,0)$ and $(1,1)$, respectively.  Then once the Cartesian coordinates of points G, H, and I are determined, it will be game over by virtue of the Law of Cosines.

Point G is uniquely determined by the intersection of the two semicircles.  Further, by symmetry, point G must have Cartesian coordinates of
$$(1/2,1/2).$$

Point H is uniquely determined by the intersection of line NB with a circular arc.  The equation for Line NB is
$$2x + y = 1.$$
The pertinent circular arc has center $(1,0)$ and radius $(1)$.  Therefore, it's equation is given by
$$(x-1)^2 + y^2 = 1.$$
Point H, which represents the intersection has coordinates
$$(1/5, 3/5).$$

Point I is uniquely determined by the intersection of line MB with a circular arc.  The equation for Line MB is
$$x + 2y = 2.$$
The same circular arc that determined point H is also pertinent for point I.  Again, it's equation is given by
$$(x-1)^2 + y^2 = 1.$$
Point I, which represents the intersection has coordinates
$$(2/5, 4/5).$$

So, a triangle is formed by the points $(1/2,1/2), (1/5,3/5),$ and $(2/5, 4/5)$.  The next step is to calculate the lengths of the sides of the triangle formed by these 3 points.
The 3 sides have lengths
$$\sqrt{\frac{1}{10}}, \sqrt{\frac{1}{10}}, \sqrt{\frac{8}{100}},$$
with the 3rd side above being across from $\angle x.$
By the Law of Cosines, this implies that
$$\cos(x) = \frac{\frac{1}{10} + \frac{1}{10} - \frac{8}{100}}{\frac{2}{10}} ~=~ \frac{3}{5}.$$
Therefore, $x = \text{Arccos}\left(\frac{3}{5}\right).$
If you look at the small diagram, on the right hand side of the original posting, you are supposed to assume that
$$\cos\left(\frac{53^{\circ}}{2}\right) \approx \frac{2}{\sqrt{5}}.$$
Using the formula that $\cos(2\theta) = 2\cos^2(\theta) - 1$, this implies that
$$\cos(53^{\circ}) \approx \frac{3}{5}.$$
Therefore, $x \approx 53^{\circ}.$
A: The two semicircles on $AD$ and $BC$ meet at the centre of the square, point $G$. Drop a perpendicular on $BM$ from point $D$ and let the foot of the perpendicular be $Q$.
Observe that, $\triangle IDG\sim \triangle BDI$ because $\frac {ID}{DG}=\frac {BD}{ID}$ and they both share $\angle IDG$. Thereafter, $\angle DIM=\frac {x}{2}$(The figure is symmetric with respect to $BD$).
$DQ=\frac {2}{\sqrt{5}}$ ($\triangle BCM\sim \triangle DQM$).
From $\triangle IDQ$,  $\cos(\frac {x}{2})=\frac {2}{\sqrt{5}}$ and from here applying the double angle identity will give $\cos(x)=\frac {3}{5} \Rightarrow x \approx \boxed {53^{\circ}}$.
A: 
Hints: As shown in figure connect A to E and extend to touch BC at I. Connect I to G and extend to touch AD at K. We have:
$BI=\frac 13 BC\Rightarrow tan (\angle BAI)=\frac 13\Rightarrow \angle BAI\approx 18.4^o$
Triangle AIK is isosceles  and $AK=2BI$, so
$\angle AIK\approx2\times 18.4\approx36.8^o$
$\angle BIA\approx90-18.4\approx71.6$
$\Rightarrow \angle BIG\approx71.6+36.8\approx108^o$
finally:
$x=\angle FGO \approx360-(2\times108+90)\approx54^o $
So all we need is to prove $BI=\frac 13 BC$. I am trying this, you can also try.
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