Walking on the edges of a rectangle

      From X-post.

       

          We will post a solution with Analytical Geometry later.

ANSWER CiP

$$\alpha=45^{\circ}$$

                    

                           Solution CiP

          Let's make some notations on the figure: $OA=a,\;OB=b,\;OC=DE=c.$

The extension of the segment $OE$ beyond point $E$ intersects the line $AD$ at point $G$.

          From the similar triangles $\Delta OBE \sim \Delta GDE$ (because $GD \parallel BO$)

 we have $\frac{OB}{GD}=\frac{EB}{ED}\;\Leftrightarrow \frac{b}{GD}=\frac{a-c}{c}$, hence

$$GD=\frac{b\cdot c}{a-c} \tag{1}$$

For the same reason, we have the similar triangles $\Delta CFO \sim \Delta DFG$, so

$$\frac{FO}{FG}=\frac{CO}{GD}\underset{(1)}{=}\frac{c}{\frac{bc}{a-c}}=\frac{a-c}{b}. \tag{2}$$

          On the other hand $\frac{AO}{AG}\overset{(1)}{=}\frac{a}{b+\frac{bc}{a-c}}=\frac{a}{\frac{ab}{a-c}}=\frac{a-c}{b}.$ Comparing with (2) it is seen that

$$\frac{FO}{FG}=\frac{OA}{AG}$$

so by the conversely of the Angle Bisector Theorem it follows that $AF$ is the angle bisector of angle $\measuredangle OAG=90^{\circ}$.

$\blacksquare$