Triangle $ABC$ has an incenter $I$. Let points $X$, $Y$ be located on the line segments $AB$, $AC$ respectively, so that $BX \cdot AB = IB^2$ and $CY \cdot AC = IC^2$. Given that the points $X, I, Y$ lie on a straight line, find the possible values of the measure of angle $A$.
1. Given the conditions \(BX \cdot AB = IB^2\) and \(CY \cdot AC = IC^2\), we start by analyzing the implications of these equations. From \(BX \cdot AB = IB^2\), we can write:
   \[
   \frac{BX}{IB} = \frac{IB}{AB}
   \]
   Since \(\angle IBA \equiv \angle IBX\) is common, triangles \(\triangle IBX\) and \(\triangle ABI\) are similar by the SAS (Side-Angle-Side) similarity criterion. Therefore, we have:
   \[
   \angle BIX = \angle BAI = \frac{\angle A}{2}
   \]

2. Similarly, from \(CY \cdot AC = IC^2\), we can write:
   \[
   \frac{CY}{IC} = \frac{IC}{AC}
   \]
   Since \(\angle ICA \equiv \angle ICY\) is common, triangles \(\triangle ICY\) and \(\triangle ACI\) are similar by the SAS similarity criterion. Therefore, we have:
   \[
   \angle CIY = \angle CAI = \frac{\angle A}{2}
   \]

3. Next, we consider the triangle \(\triangle BCI\). The incenter \(I\) forms the angle:
   \[
   \angle BIC = 180^\circ - \left( \frac{\angle B}{2} + \frac{\angle C}{2} \right)
   \]
   Using the fact that the sum of the angles in a triangle is \(180^\circ\), we have:
   \[
   \angle A + \angle B + \angle C = 180^\circ
   \]
   Therefore:
   \[
   \angle BIC = 180^\circ - \left( \frac{180^\circ - \angle A}{2} \right) = 180^\circ - 90^\circ + \frac{\angle A}{2} = 90^\circ + \frac{\angle A}{2}
   \]

4. Given that points \(X, I, Y\) are collinear, the angle \(\angle XIY\) must be a straight angle, i.e., \(180^\circ\). We can express \(\angle XIY\) as the sum of the angles \(\angle BIX\), \(\angle BIC\), and \(\angle CIY\):
   \[
   \angle XIY = \angle BIX + \angle BIC + \angle CIY
   \]
   Substituting the known angles, we get:
   \[
   \angle XIY = \frac{\angle A}{2} + 90^\circ + \frac{\angle A}{2} = 180^\circ
   \]
   Simplifying, we have:
   \[
   \frac{\angle A}{2} + 90^\circ + \frac{\angle A}{2} = 180^\circ
   \]
   \[
   \angle A + 90^\circ = 180^\circ
   \]
   \[
   \angle A = 90^\circ
   \]

The final answer is \(\boxed{90^\circ}\).