Determine the angles of triangle $ABC$ if we know that two of its altitudes are at least as long as the corresponding base.
Let's denote the sides and altitudes of the triangle in the usual way. Let the two altitudes be \( m_{a} \) and \( m_{b} \), that is,

\[
m_{a} \geq a, \quad m_{b} \geq b
\]

We also know that

\[
m_{a} \leq b, \quad \text{and} \quad m_{b} \leq a,
\]

since, for example, \( m_{a} \) is a leg in a right triangle where \( b \) is the hypotenuse.

From (1) and (2),

\[
a \leq m_{a} \leq b \quad \text{and} \quad b \leq m_{b} \leq a
\]

and therefore

\( a \leq b \leq a \), which is only possible if \( a = b = m_{a} = m_{b} \), meaning the triangle is isosceles and right-angled, i.e., \( \angle ACB = 90^\circ \), and \( \angle ABC = \angle BAC = 45^\circ \).