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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 $$\frac{|AP|}{\sin x}=\frac{|AB|}{\sin (\beta +\gamma )}=\frac{|AB|}{\sin \alpha }\text{and}\frac{|BP|}{\sin x}=\frac{|BC|}{\sin (\gamma +\alpha )}=\frac{|BC|}{\sin \beta },$$ 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 $$ \frac{|AC| \cdot |BP|}{|BC|} = \frac{|BC| \cdot |AP|}{|AB|}.