Finding a missing angle in the picture containing regular hexagon and square I want to find $\angle AGM=\theta$ in the following picture:

Here $ABCDEF$ and $BAGH$ are regular hexagon and square respectively and $M$ is the midpoint of $FH$.
I found a trigonometric solution. I'm providing key ideas of the solution:
Let $AB=1$. Now we can apply cosine rule on $\triangle AHF$ to find $HF$ and $HM$. Now in $\triangle MGH$, we can find $GM$ using cosine rule again and then find $\angle MGH$ by sine rule. This gives $\theta=15^{\circ}$. (I'm not providing the calculations as they are not nice and I did most of them with calculator.)
But I believe there are some beautiful synthetic solution to the but didn't find one. So, I need a synthetic solution to the problem.
 A: Let $I$ be a center of hexagon. Then $HG = ID$ and they are parallel, so $IDGH$ is a paralelogram so $K$ is also the midpoint of $GI$, thus $G,K,I$ are collinear.
Since $GEI$ is isosceles triangle and $\angle GEI = 150^{\circ}$ we have $\theta = 15^{\circ}$.

A: Even without pure geometry, the work can be simplified. Also, see my edit at the end for a synthetic solution.

$\angle PBC = \angle AFQ = 30^\circ$
If side length is $a$, $PC = GQ = \frac{a}{2}$
So, $PF = HQ = \frac{3a}{2}$
Similarly, $PH = FQ = a + \frac{a \sqrt3}{2}$
Given $M$ is the midpoint of $FH$,
$MN = a - \frac{HQ}{2} = \frac{a}{4}$
$GN = \frac{FQ}{2} = \frac{a (2 + \sqrt3)}{4}$
$\tan \theta = \frac{1}{2 + \sqrt3} = 2 - \sqrt3 \implies \theta = 15^0$

Synthetic solution (using similar construct as above):

Given $M$ is the midpoint of $HF$, it is also the center of rectangle $HPFQ$ and hence of rectangle $GIJK$.
Also note $J$ is the center of the hexagon.
So, $\triangle FAJ$ is an equilateral triangle.
$\angle JAK = 30^\circ$.
In $\triangle GAJ$, $AG = AJ$
$\therefore \angle AGM = \angle AJM = 15^\circ$
