One, (25 points) In trapezoid $ABCD$, $AB \parallel CD, AB=125, CD=$ $DA=80$. Can diagonal $BD$ divide the trapezoid into two similar triangles? If not, provide a proof; if so, find the lengths of $BC, BD$.
First, assume that two triangles can be similar, and $AD$ is the corresponding side.
(1) If $AD$ and $DC$ are corresponding sides, then the similarity ratio is 1, and the two triangles are congruent. Their corresponding angles $\angle ABD = \angle DBC$, but from $AB \parallel CD$, we know $\angle ABD = \angle BDC$, so $\angle CBD = \angle CDB$. In this case, $DB$ can be either 80 or 125. Both scenarios contradict the congruence of the two triangles, so $AD$ and $DC$ cannot be corresponding sides.
(2) If $AD$ and $BD$ are corresponding sides, then $\angle ABD = \angle BCD$, but $\angle ABD = \angle BDC$, so $\angle BCD = \angle BDC$. $\triangle CDB$ is an isosceles triangle, and thus $\triangle ABD$ is also an isosceles triangle. $DB$ can be either 80 or 125. If $BD = 80$, then $\triangle BCD$ is an equilateral triangle, which cannot be similar to $\triangle ABD$. If $BD = 125$, then $BC = 125$, which contradicts the correspondence of $AD$ and $BD$.
(3) If $AD$ and $BC$ are corresponding sides, then only
$$
\frac{BC}{AD} = \frac{CD}{BD} = \frac{BD}{AB},
$$

i.e., $\frac{BC}{80} = \frac{80}{BD} = \frac{BD}{125}$,
we have $BD = \sqrt{80 \times 125} = 100$,
$$
BC = \frac{80^2}{BD} = 64.
$$

In summary, the diagonal $BD$ can divide the trapezoid into two similar triangles, and in this case, there is a unique solution $BC = 64, BD = 100$.
Note: If we directly consider $AD$ and $BC$ as corresponding sides, it cannot be confirmed.