Is it you Thales?

Let the radius of the semicircle be 1, then we note that $BE=1$ and $BF=$ $EG=$ $EF=$ $BG =$ $\sin 45^\circ =$ $\frac 1{\sqrt 2}$ . By sine rule , we have

$\begin{aligned} \frac {\sin \angle BDG}{BG} & = \frac {\sin \angle DGB}{BD} \\ \frac {\sin \angle BDG}{\frac 1{\sqrt 2}} & = \frac {\sin 45^\circ}{1} = \frac 1{\sqrt 2} \\ \sin \angle BDG & = \frac 12 \\ \implies \angle BDG & = 30^\circ \end{aligned}$

$\implies \angle DBA = \angle BDG + \angle DGB = 30^\circ + 45^\circ = 75^\circ$ and $\angle DAC = \dfrac {180^\circ - \angle DBA}2 = \dfrac {180^\circ - 75^\circ}2 = \boxed{52.5}^\circ$ .