Problem 1065

1, Proof: As shown in the figure, connect $\mathrm{AD}$ intersecting circle $\mathrm{P}$ at $\mathrm{S}$. We know that
$\angle U A D = \angle E A S = \angle B C D$. It is also easy to see that $A, U, D, V$ are concyclic, so $\angle U A D = \angle U V D = 90^{\circ} - \angle A V U = 90^{\circ} - \angle A F E$. Therefore, $\angle E A S + \angle A E F = 90^{\circ}$, which means $AS$ is the diameter of circle $P$, i.e., $A, P, S, D$ are collinear.
Let $\mathrm{TK}$ intersect circle $\mathrm{P}$ at $\mathrm{L}$, then $\angle T L S = \angle T A S = \angle T K D$, so $S L \parallel D K \parallel E F$, hence $\triangle S E F \cong \triangle L F E$.

Also, $\angle S E F = \angle D A F = \angle D B C$, $\angle S F E = \angle D A B = \angle D C B$, so $\triangle S E F \sim \triangle D B C$. Take point $\mathrm{Q}$ such that
$\angle Q E F = \angle N B C$, $\angle Q F E = \angle N C B$, then quadrilateral $\mathrm{QESF}$ is similar to quadrilateral $\mathrm{NBDC}$. By the similarity correspondence,
we only need to prove that $\mathrm{P}, \mathrm{Q}, \mathrm{~S}$ are collinear. Note that $\mathrm{M}, \mathrm{N}$ are a pair of isogonal conjugates, so
$\angle Q E F = \angle M B E$, $\angle Q F E = \angle M C F$, thus $\angle Q E A = \angle B M E$, $\angle Q F A = \angle C M F$.
According to the trigonometric form of Ceva's theorem, to prove that $\mathrm{P}, \mathrm{Q}, \mathrm{~S}$ are collinear, we only need to prove:
$$\begin{aligned} & \frac{\sin \angle Q E F}{\sin \angle A E Q} \times \frac{\sin \angle E A S}{\sin \angle F A S} \times \frac{\sin \angle Q F A}{\sin \angle Q F E} = 1 \\ \Leftrightarrow & \frac{\sin \angle M B E}{\sin \angle B M E} \times \frac{\sin \angle E A S}{\sin \angle F A S} \times \frac{\sin \angle C M F}{\sin \angle M C F} = 1 \\ \Leftrightarrow & \frac{M E}{B E} \times \frac{E S}{F S} \times \frac{C F}{M F} = 1 \\ \Leftrightarrow & \frac{M E}{B E} \times \frac{F L}{E L} \times \frac{C F}{M F} = 1 \end{aligned}$$

In fact, note that $\angle T B E = \angle T C F$, $\angle T E A = \angle T F A$, then $\triangle T B E \sim \triangle T C F$, so $\frac{M E}{M F} = \frac{E L \times T E}{F L \times T F} = \frac{E L \times B E}{F L \times C F} \Rightarrow \frac{M E}{B E} \times \frac{F L}{E L} \times \frac{C F}{M F} = 1$, hence the proposition is proved.