Circle Inscribed in a Half of a SQUARE

 From a Linkedin Post

ANSWER CiP

$$\frac{a}{b}=1+\sqrt{2}$$

                    Solution CiP

          Notations

We have the equality of triangles (ASA)

$$\Delta BEO \equiv \Delta DFO$$

because $OB \equiv OD:=a$, $\measuredangle {OBE}=\measuredangle{ODF}=45^{\circ}, \; \measuredangle {BOE} \equiv \measuredangle{DOF}$ as vertical angles.

From here $DF=b$, so $CF=a$. But $CF \equiv CO$ as tangents from point C to the circle. Then, from the right and isosceles triangle $BOC$ we find $BC=a\cdot \sqrt{2}$.

          We have the equations $a+b=AB=BC=a\cdot \sqrt{2}$, hence

 $a(\sqrt{2}-1)=b\;\Rightarrow a(\sqrt{2}-1)(\sqrt{2}+1)=b(\sqrt{2}+1) \Leftrightarrow a(2-1)=b(\sqrt{2}+1)$, so $\frac{a}{b}=\sqrt{2}+1$. We got the answer.

$\blacksquare$