A half-circle is inscribed in a square such that its diameter is the side length of the square

Question

A tangent is drawn from a vertex of a square to the half-circle and intersects one of the side lenghts of the square in a point. What is the ratio that that point divides the square length into? Or in other words, what is the ratio of $DH$ to $HC$ in the picture?

I have tried proving that the triangle $FGC$ is equilateral, but to no success.

Answer 1

$$\measuredangle AHD=\measuredangle BAH=2\arctan\frac{1}{2}.$$ Thus, $$DH=AD\cot\left(2\arctan\frac{1}{2}\right)=AD\cdot\frac{1-\left(\frac{1}{2}\right)^2}{2\cdot\frac{1}{2}}=\frac{3}{4}DC.$$ Thus, $$DH:HC=3:1.$$

Answer 2

We have a few right triangles $ABF\cong AGF$ and $FCH\cong FGH$. Then $\angle CFH=\frac\pi2-\angle AFB$, hence $\angle CFH=\angle BAF$, i.e., all four right triangles are similar. Then $HC:FC=BF:AB=1:2$ and ultimately $HC=\frac14DC$.