Croatian Mathematical Competitions

Let us denote $\angle BAC = \alpha$, $\angle CBA = \beta$ and $\angle ACB = \gamma$.

From $\angle APB = \beta + \gamma$ and $\angle BPC = \gamma + \alpha$ we get
$$\angle CPA = 360^\circ - \angle APB - \angle BPC = 360^\circ - \beta - \gamma - \alpha = \alpha + \beta.$$
If $\angle PBA = x$, then $\angle BAP = 180^\circ - \angle APB - \angle PBA = 180^\circ - \beta - \gamma - x = \alpha - x$, so $\angle PAC = \angle BAC - \angle BAP = \alpha - (\alpha - x) = x$.
Analogously we get $\angle ACP = \gamma - x$, $\angle PCB = x$ and $\angle CBP = \beta - x$.
Applying the sine rule to the triangles $ABP$ and $BCP$ we get
$$\frac{|AP|}{\sin x} = \frac{|AB|}{\sin (\beta + \gamma)} = \frac{|AB|}{\sin \alpha} \quad \text{and} \quad \frac{|BP|}{\sin x} = \frac{|BC|}{\sin (\gamma + \alpha)} = \frac{|BC|}{\sin \beta},$$
wherefrom due to $\sin \alpha : \sin \beta = |BC| : |AC|$ we get
$$\frac{|BP|}{|AP|} = \frac{|BC| \sin \alpha}{|AB| \sin \beta} = \frac{|BC| \cdot |BC|}{|AB| \cdot |AC|},$$
which finally leads to
$$$$
$\text{|AC|}$ |BP|}{|BC|} = $\text{|BC|}$ |AP|}{|AB|}.