Where do you see lots of Asians?

Since $BD = FE || BC$ , $D$ lies on $BC$ , and $F$ lies on $AB$ , you can deduce that $BDEF$ is a rectangle snugly against $AB$ and $BC$ .

$FGE$ and $BDC$ are both straight lines, and $BF$ , $DG$ and $CE$ all meet at $A$ . From this, we deduce that $G$ splits $FE$ in the same ratio as $D$ splits $BC$ . In other words, $\frac{FE}{FG} = \frac{BC}{BD}.$ But $FE = BD$ by construction, therefore $BC.FG = (FE)^2.$

If we let $AB = 1$ , then $FE = 1$ by construction, and $BC = \tan 65^\circ$ . Then we have $FG = \cot 65^\circ = \tan 25^\circ.$ But $BHGF$ is a rectangle, therefore $FG = BH$ . So $BH = \tan 25^\circ \\ \Rightarrow \frac{BH}{BA} = \tan 25^\circ \\ \Rightarrow \angle BAH = \boxed{25^\circ}.$

(Coordinate Geometry FTW)

(Please draw a diagram to comprehend this)

Let B be origin, A be (0,1), C be ( $\tan65^\circ$ ,0).

Then, eqn. of AC: $(\tan65^\circ)y+x=\tan65^\circ$ .

D is (1,0), BDEF is a rectangle, so x-coordinate of E is 1.

E lies on AC, so sub. x=1 to eqn. of AC to get $y=\dfrac{\tan65^\circ-1}{\tan65^\circ}$ .

That is the y-coordinate of E.

Eqn. of AD: $x+y=1$ .

Eqn. of EF: $y=\dfrac{\tan65^\circ-1}{\tan65^\circ}$ .

Solve, $x=\dfrac1{\tan65^\circ}=\tan25^\circ$ .

Therefore, H is ( $\tan25^\circ$ ,0).

BAH is a right-angled triangle, and $\tan\angle BAH=\dfrac{\tan25^\circ}1$ .

Therefore, $\angle BAH=25^\circ$ .