Since $\angle ADN+\angle ADM=\angle ACB+\angle ABC=180^{\circ}-\angle MAN,$ then $AMDN$ is cyclic, being $D$ the midpoint of its circumcircle arc $MN.$ Hence $\angle ADN=\angle AMN$ and $\angle DAN=\angle DAM$ gives $\triangle ADN \sim \triangle AMP$ $\Longrightarrow$ $AM \cdot AN=AP \cdot AD \ (1).$ But $\angle MDA=\angle ABC$ $\Longrightarrow$ $AD^2=AM \cdot AB$ and similarly $AD^2=AN \cdot AC,$ hence $AD^4=AB \cdot AC \cdot AM \cdot AN \ (2).$ Combining $(1)$ and $(2)$ yields $AD^3=AB \cdot AC \cdot AP.$

Lemma 1. $BMNC$ is cyclic. Proof. Since $AD$ bisects $\angle BAC$, $\triangle{AMD} \sim \triangle{ADB}$ and $\triangle{ADN} \sim \triangle{ACD}$. Then $AD^2=AM \cdot AB$ $(1)$ $AD^2=AN \cdot AC$ Then, $AM \cdot AB = AN \cdot AC$. So, we proved. In other words, by Lemma 1, $\angle{ACD}=\angle{AMP}$ $\Rightarrow$ $\triangle AMP \sim \triangle ACD $ $\Rightarrow$ $AD \cdot AM = AP \cdot AC$ $(2)$ Then, multiplying $(1)$ and $(2)$: $AD^3=AP \cdot AB \cdot AC $. So we are done!

this was also 2000 Brazil Cono Sur TST 2.3