15. Given that the diameter of the circumcircle of $\triangle A B C$ is 25, the lengths of the three sides are all integers, and the distances from the circumcenter $O$ to the sides $A B$ and $B C$ are also integers, with $A B > B C$. Find the lengths of the three sides of $\triangle A B C$.
15. As shown in Figure 3, draw perpendiculars from the center $O$ to $AB$ and $BC$, with the feet of the perpendiculars being $D$ and $E$ respectively. Let $AB = x, OD = y$ (where $x, y$ are positive integers). Then
$BD = \frac{x}{2}$.
Since $DB^2 + OD^2 = OB^2$, we have $x^2 + 4y^2 = 25^2$.
Solving this, we get $(x, y) = (7, 12)$ or $(15, 10)$, which means $AB = 7, OD = 12$ or $AB = 15, OD = 10$.
Similarly, $BC = 7, OE = 12$ or $BC = 15, OE = 10$.
Since $AB > BC$, we have
$$
AB = 15, OD = 10, BC = 7, OE = 12.
$$

Since $OD \perp AB, OE \perp BC$, points $O, D, B, E$ are concyclic.
By Ptolemy's theorem,
$$
DE \cdot OB = OD \cdot BE + OE \cdot DB,
$$

which gives $DE = \frac{OD \cdot BE + OE \cdot DB}{OB} = 10$.
Since point $D$ bisects $AB$ and point $E$ bisects $BC$, $DE$ is a midline.

Thus, $AC = 20$, meaning the lengths of the sides of $\triangle ABC$ are $15, 7, 20$.