Problem 1417

(1) Denote $\angle NPC = \phi_1$, $\angle MPC = \phi_2$, then
$$\frac{S_{\triangle NPC}}{S_{\triangle NPA}} = \frac{CN}{AN} = \frac{CP \sin \phi_1}{AP \cos \phi_2}.$$
Therefore, $\tan \phi_1 = \frac{CN}{AN} \cdot \frac{AP}{CP}$.
Similarly, $\tan \phi_2 = \frac{CM}{BM} \cdot \frac{BP}{CP}$.
Thus, $\tan \phi_1 = \tan \phi_2$ is equivalent to
$$\frac{CN}{AN} \cdot \frac{AP}{BP} \cdot \frac{BM}{CM} = 1.$$
In $\triangle ABC$, applying Ceva's theorem to the lines $AM, BN, CP$ gives the above equation directly.

(2) Denote $\angle NPC = \angle MPC = \phi$.
To prove the conclusion, it suffices to prove
$$\frac{\sin(\phi - x)}{\sin x} = \frac{\sin(\phi - y)}{\sin \gamma},$$
where $x = \angle EPO, \gamma = \angle DPO$. In fact, the latter is equivalent to
$$\begin{align*} & \frac{\frac{\sin \phi \cos x - \cos \phi \sin x}{\sin x}}{\sin \gamma} \\ &= \frac{\sin \phi \cos x - \cos \phi \sin \gamma}{\sin \gamma} \\ \Leftrightarrow \quad & \cot x = \cot \gamma \Leftrightarrow x = \gamma. \end{align*}$$
Suppose $E \in NH, D \in CM$, then
$$\frac{NE}{EH} = \frac{S_{\triangle NEP}}{S_{\triangle EHP}} = \frac{NP \cdot \sin(\phi - x)}{PH \cdot \sin x}.$$
$$Therefore, \frac{\sin(\phi - x)}{\sin x} = \frac{NE}{EH} \cdot \frac{PH}{NP} \quad (1)$$
Similarly, we obtain
$$\frac{\sin(\phi - \gamma)}{\sin \gamma} = \frac{DM}{CD} \cdot \frac{CP}{PM} \quad (2)$$
Using (1) and (2), it suffices to prove
$$\frac{NE}{EH} \cdot \frac{CD}{DM} \cdot \frac{PH}{CP} \cdot \frac{MO}{NO} = 1. \quad (3)$$
(since $PO$ is the angle bisector of $\Delta NPM$, that is, $\frac{PM}{PN} = \frac{MO}{NO}$.)
Also, since
$$\frac{NE}{EH} = \frac{S_{\Delta NEO}}{S_{\Delta EHO}} = \frac{NO \sin \delta}{OH \sin \psi}, \\ \frac{CD}{DM} = \frac{S_{\Delta CDO}}{S_{\Delta DMO}} = \frac{CO \sin \psi}{OM \sin \delta},$$
where $\delta = \angle MOD, \psi = \angle EOP$. Thus, (3) simplifies to
$$\frac{OC}{OH} \cdot \frac{PH}{PC} = 1.$$
Applying Menelaus's theorem to $\Delta BHC$ and line $MN$, to $\Delta CHM$ and line $AB$, and to $\Delta BHM$ and line $AC$ respectively, we obtain
$$\frac{BN}{NH} \cdot \frac{HO}{OC} \cdot \frac{CM}{MB} = 1, \\ \frac{CP}{PH} \cdot \frac{HA}{AM} \cdot \frac{MB}{BC} = 1, \\ \frac{HN}{NB} \cdot \frac{BC}{CM} \cdot \frac{MA}{HA} = 1,$$
Multiplying the three equations gives
$$\frac{OC}{OH} \cdot \frac{PH}{PC} = 1.$$