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?
Best Answer
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 $\angle HGQ \approx 26.5^\circ$
$\therefore x \approx 53^\circ$