Solution. From the similarity of triangles $ABO$ and $CDO$, it follows that $\overline{CO}: \overline{AO} = \overline{CD}: \overline{AB}$, and from this

$$(\overline{CO} + \overline{AO}): \overline{AO} = (\overline{CD} + \overline{AB}): \overline{AB}$$

or equivalently

$$\overline{AC}: \overline{AO} = \overline{AC}: \overline{AO}$$

so $\overline{AO} = \overline{AB}$. Therefore, triangle $ABO$ is equilateral, and similarly, $\triangle CDO$ is equilateral. This means that $BP \perp AO$ (where $P$ is the midpoint of $\overline{AO}$), so triangle $BCP$ is right-angled, with hypotenuse $BC$. Since $PQ$ is the median to the hypotenuse $BC$, it follows that

$$\overline{PQ} = \frac{1}{2} \overline{BC}$$

Similarly, from the right-angled triangle BCR, we have

$$\overline{RQ} = \frac{1}{2} \overline{BC}$$

Furthermore, in triangle ADO, PR is the midline, so

$$\overline{PM} = \frac{1}{2} \overline{AD}$$

!

Finally, considering the condition of the problem $\overline{AO} = \overline{AB}$ and (1), (2), and (3), we conclude that triangle $PQR$ is equilateral, with a side equal to half the leg $\overline{AD}$.