一、Fill in the Blanks (8 points per question, total 64 points)
1. In an acute triangle $\triangle A B C$, the angle bisector $A D$, median $B M$, and altitude $C H$ intersect at a point $P$. Then $C H$ $\qquad$ $A H$ (fill in “ $>$ ”, “ $<$ ”, or “ $=$ ”).
- 1. $>$.

From Ceva's Theorem, as shown in Figure 2, we have
$$
\begin{array}{l}
\frac{A M}{M C} \cdot \frac{C D}{D B} \cdot \frac{B H}{H A}=1 . \\
\text { Since } A M=M C, \\
\frac{C D}{D B}=\frac{A C}{A B}=\frac{b}{c}, \\
\frac{B H}{H A}=\frac{B C \cos B}{A C \cos A} \\
=\frac{a \cos B}{b \cos A},
\end{array}
$$

Therefore, $a \cos B=c \cos A$.
Draw $B E \perp A C$ at point $E$, intersecting $C H$ at point $Q$. Then $B H=A E$. Thus, $B E \geqslant B Q>B H=A E$.
Hence $\angle B A C>45^{\circ}$. Therefore, $C H>A H$.