5. In triangle $A B C$, $A C>A B$. A height $A H$ is dropped from vertex $A$. Which difference is greater, $(A C-A B)$ or $(H C-H B)$?
Solution. We need to consider two cases, see the figure.
![](https://cdn.mathpix.com/cropped/2024_05_06_5c672c8ae4324133d019g-3.jpg?height=328&width=1376&top_left_y=384&top_left_x=407)

1st case. Point $H$ lies inside the segment $B C$. By the Pythagorean theorem,

$$
A B^{2}-B H^{2}=A H^{2}=A C^{2}-C H^{2}
$$

We transition to the difference of squares and the comparison of segments.

$$
\begin{gathered}
C H^{2}-B H^{2}=A C^{2}-A B^{2} \\
(C H-B H)(C H+B H)=(A C-A B)(A C+A B) \\
(C H-B H) \cdot B C=(A C-A B)(A C+A B)
\end{gathered}
$$

By the triangle inequality, $A C+A B>B C$. Since the products are equal, and the second factor in the right product is greater than the second factor in the left product, then $C H-B H>A C-A B$.

2nd case. Point $H$ lies outside the segment $B C$ or coincides with point $B$. By the Pythagorean theorem,

$$
A B^{2}-B H^{2}=A H^{2}=A C^{2}-C H^{2}
$$

From this equation and the condition $A C>A B$, it follows that $B$ lies between $H$ and $C$, i.e., $C H>B H$. (In particular, $B H=0$, $H$ coincides with point $B$). Therefore, $C H-B H=B C$, and we need to compare $A C-A B$ and $B C$. By the triangle inequality, $B C+A B>A C$, so $B C>A C-A B$.

We have obtained that in both cases $C H-B H>A C-A B$.

Answer. $C H-B H>A C-A B$.