*Answer:* $45^\circ$, $90^\circ$, $135^\circ$.

Let $\alpha = \angle BAC$. Both triangles $BEA$ and $BEC$ have right angle at vertex $E$. Hence these triangles can be isosceles only if their other angles have size $45^\circ$.

Suppose that $BEC$ is isosceles (see figure below).
Then $\angle BCE = 45^\circ$. Since $\angle BCE = \angle BCA = \frac{180^\circ - \alpha}{2}$, we have $\alpha = 180^\circ - 2 \cdot 45^\circ = 90^\circ$.

Suppose now that $BEA$ is isosceles. Then $\angle BAE = 45^\circ$.
If $E$ lies on the line segment $AC$ (see figure below) then $\angle BAE = \angle BAC = \alpha$, implying $\alpha = 45^\circ$.
If $E$ lies outside the line segment $AC$ (see figure below) then $\angle BAE = 180^\circ - \angle BAC = 180^\circ - \alpha$, implying $\alpha = 180^\circ - 45^\circ = 135^\circ$.

Let $\alpha = \angle BAC$. Then $\angle ABC = \angle ACB = \frac{180^\circ - \alpha}{2}$ and $\angle EBC = 90^\circ - \frac{180^\circ - \alpha}{2} = \frac{\alpha}{2}$.
Both triangles $BEA$ and $BEC$ have right angle at vertex $E$. Hence these triangles can be isosceles only if their other two angles have equal size.

Suppose that $BEC$ is isosceles. Then $\frac{180^\circ - \alpha}{2} = \frac{\alpha}{2}$, implying $\alpha = 90^\circ$.

Suppose now that $BEA$ is isosceles. If the triangle $ABC$ is acute, we must have $\angle BAE = \alpha$ and $\angle ABE = \frac{180^\circ - \alpha}{2} - \frac{\alpha}{2} = 90^\circ - \alpha$. Thus $\alpha = 90^\circ - \alpha$, implying $\alpha = 45^\circ$.
If the triangle $ABC$ is obtuse then $\angle BAE = 180^\circ - \alpha$ and $\angle ABE = \frac{\alpha}{2} - \frac{180^\circ - \alpha}{2} = \alpha - 90^\circ$. Hence $180^\circ - \alpha = \alpha - 90^\circ$, implying $\alpha = 135^\circ$.
The triangle $ABC$ is not right as, otherwise, the triangle $BEA$ would have two right angles.