$ABCD$ is a square with an inscribed semicircle and tangent $DT$, find the measure of $\frac{\alpha}{\theta}$ As title suggests, in the figure given below with a square, a semicircle and some unknown angles $\alpha$ and $\theta$, the goal is to find the numerical value of the expression: $$\frac{\alpha}{\theta}$$
I'm not entirely sure how to go about this problem, its a fairly unique type of problem that I've never encountered before. Nevertheless, I'll also post my own approach below, please share your own as well!

 A: I will add the picture. My proof without words:

A: This is my approach. I'll add an explanation below as well:

Here's my explanation:
1.) Connect point $C$ and $A$ via the diagonal $AC$. Next, also, connect the tangency point $T$ with the point of intersection of both diagonals (let's call it point $X$). Lastly, connect the tangency point $T$ with $A$. Using the properties of cyclic quadrilaterals and some circle theorems, we can deduce that $\angle BTA=\angle BXA=90$. And also that $\angle TBX=\angle TAX=\alpha$.
2.) Notice that since $DT$ is tangent to the semicircle, it is equal to segments $DA$ and $DC$ (all the sides of the square). This means that via the inscribed angle theorem, $\angle TDC$ is twice the measure of $\angle TAX$. Thus $\angle TDC=2\alpha$. This also implies that $\angle TDA=90-2\alpha$. Once again, via the inscribed angle theorem, this means that $\angle TCA$ is half of $\angle TDA$, thus $\angle TCA=45-\alpha$. However, since $AC$ is a diagonal and $\angle TCB=\theta$, we also know that $\angle TCA=45-\theta$. This implies that:
$$45-\alpha=45-\theta$$
and thus:
$$\frac{\alpha}{\theta}=1$$
