Three, (25 points) It is known that the lengths of the three sides of the acute triangle $\wedge A B C$ and the length of one of its altitudes are four consecutive integers, and this altitude divides $\triangle A B C$ into two right triangles with integer side lengths. Try to find the lengths of the three sides of $\triangle A B C$.
Three, as shown in Figure 7, let the height be $A D$, and the four consecutive integers be $n, n+1, n+2, n+3$. Also, assume $A B < A C$, then $A D < A B < A C$. Therefore, $A D = n$ or $n+1$.

When $A D = n+1$, only $B C = n$. Thus, $A B = n+2, A C = n+3$. So, $B D = \sqrt{2 n+3}, C D = 2 \sqrt{n+2}$, and both are integers. By $B C = B D + C D$, we get
$$
n = \sqrt{2 n+3} + 2 \sqrt{n+2} \text{. }
$$

This equation has no integer solutions.
When $A D = n$, $B D \leqslant \sqrt{(n+2)^{2}-n^{2}} = 2 \sqrt{n+1}$, $C D \leqslant \sqrt{(n+3)^{2}-n^{2}} = \sqrt{6 n+9}$.
$$
\therefore n+1 \leqslant B C \leqslant 2 \sqrt{n+1} + \sqrt{6 n+9} < 5 \sqrt{n+1} \text{. }
$$

Solving this, we get $n < 24$.
(1) If $B C = n+1$, then $A B = n+2, A C = n+3$. In this case, $B D = 2 \sqrt{n+1}, C D = \sqrt{6 n+9}$. There is no $n$ that satisfies $n < 24$ and both $2 \sqrt{n+1}$ and $\sqrt{6 n+9}$ are integers.
(2) If $B C = n+2$, then $A B = n+1, A C = n+3$. In this case, $B D = \sqrt{2 n+1}, C D = \sqrt{6 n+9}$. The only $n$ that satisfies $n < 24$ and both $\sqrt{2 n+1}$ and $\sqrt{6 n+9}$ are integers is $n = 12$. Thus, $A B = 13, B C = 14, A C = 15$.
(3) If $B C = n+3$, then $A B = n+1, A C = n+2$. In this case, $B D = \sqrt{2 n+1}, C D = 2 \sqrt{n+1}$. There is no $n$ that satisfies $n < 24$ and both $\sqrt{2 n+1}$ and $2 \sqrt{n+1}$ are integers.

In summary, the triangle $\triangle A B C$ that meets the conditions has only $\cdots$ one, and the side lengths are $13, 14, 15$.