Right triangle inscribed in a square. Find the square area?

Without using trigonometry:

enjoy...

By coloring triangles the same I am trying to emphasize the similar triangles, which are not necessarily equal (although red and blue triangles are). I don't mean that the triangles with the same color has the same area, be careful.

I think your way is good, but we don't need to find $\tan\theta$.

From $4\sin\theta=4\cos\theta+3\sin\theta$, we have $$\sin\theta=4\cos\theta$$ Squaring the both sides gives $$\sin^2\theta=16(1-\sin^2\theta)$$ from which we can have $$\sin^2\theta=\frac{16}{17}\quad\Rightarrow\quad \text{(area)}=16\sin^2\theta=\frac{16^2}{17}$$

Here it's another way. The triangles AhC and Bhf are similar. If you put $Bh=x$ and $Bf=y$ you get the relations $$x+\frac{4}{3}y=\frac{4} {3} x$$ $$9=x^2+y^2$$ From which you can obtain the length of the side and then the area