CroatianCompetitions

Let us denote $\angle ABP = \angle BCP = \varphi$ and $\angle PBC = \angle PCA = \psi$.
Let $T$ be the midpoint of the segment $AC$. Since $AC$ is the hypotenuse of the right triangle $APC$, it follows that $|AT| = |PT| = |CT|$. Hence $\angle CPT = \psi$.

Since $\angle DPC$ is the exterior angle of the triangle $BCP$, we have
$$\angle DPC = \angle PCB + \angle PBC = \varphi + \psi$$
so $\angle DPT = \varphi$.

Applying the law of sines on triangles $ABD$ and $PDT$ we get
$$\frac{|AD|}{|AB|} = \frac{\sin \angle ABD}{\sin \angle ADB}, \quad \frac{|DT|}{|PT|} = \frac{\sin \angle DPT}{\sin \angle PDT}$$
and then
$$\frac{|AD|}{|AB|} = \frac{\sin \angle ABD}{\sin \angle ADB} = \frac{\sin \varphi}{\sin (180^\circ - \angle ADB)} = \frac{\sin \angle DPT}{\sin \angle PDT} = \frac{|DT|}{|PT|}$$
Now it follows that
$$\frac{|AD|}{|AC|} = \frac{|AD|}{|AB|} = \frac{|DT|}{|PT|} = \frac{|AT| - |AD|}{|PT|} = \frac{\frac{1}{2}|AC| - |AD|}{\frac{1}{2}|AC|} = 1 - 2 \cdot \frac{|AD|}{|AC|}$$
so $|AD| : |AC| = 1 : 3$. The desired ratio is $|AD| : |CD| = 1 : 2$.