Let $ABC$ be a triangle with angles $\alpha$, $\beta$, and $\gamma$. Let $X$, $Y$, and $Z$ be the points where the angle bisectors of angles $A$, $B$, and $C$ intersect the circumcircle of triangle $ABC$, respectively. We want to prove that the angles of triangle $XYZ$ are $90^{\circ} - \frac{\alpha}{2}$, $90^{\circ} - \frac{\beta}{2}$, and $90^{\circ} - \frac{\gamma}{2}$.

1. Problem Description

Let $ABC$ be a triangle with angles $\alpha$, $\beta$, and $\gamma$. Let $X$, $Y$, and $Z$ be the points where the angle bisectors of angles $A$, $B$, and $C$ intersect the circumcircle of triangle $ABC$, respectively. We want to prove that the angles of triangle $XYZ$ are $90^{\circ} - \frac{\alpha}{2}$, $90^{\circ} - \frac{\beta}{2}$, and $90^{\circ} - \frac{\gamma}{2}$.

2. Solution Steps

Let $I$ be the incenter of triangle $ABC$.

Since $AX$ is the angle bisector of angle $A$, we have $\angle BAX = \angle CAX = \frac{\alpha}{2}$.

Similarly, $\angle ABY = \angle CBY = \frac{\beta}{2}$ and $\angle ACZ = \angle BCZ = \frac{\gamma}{2}$.

We know that $\angle BAC + \angle ABC + \angle ACB = \alpha + \beta + \gamma = 180^{\circ}$.

Consider $\angle YXZ$. This angle is subtended by the arc $YZ$. Thus, $\angle YXZ = \angle YAZ + \angle YCZ$.

We have $\angle YAZ = \angle YAC = \angle YBC = \frac{\beta}{2}$ since $BY$ is the angle bisector of $\angle B$.

We also have $\angle ZAC = \angle ZBC = \frac{\gamma}{2}$ since $CZ$ is the angle bisector of $\angle C$.

Therefore,

$$\angle YXZ = \frac{\beta}{2} + \frac{\gamma}{2} = \frac{\beta + \gamma}{2}$$

Since $\alpha + \beta + \gamma = 180^{\circ}$, we have $\beta + \gamma = 180^{\circ} - \alpha$.

$$\angle YXZ = \frac{180^{\circ} - \alpha}{2} = 90^{\circ} - \frac{\alpha}{2}$$

Similarly,

$\angle XZY = \angle XAY + \angle XBY = \frac{\alpha}{2} + \frac{\beta}{2} = \frac{\alpha + \beta}{2} = \frac{180^{\circ} - \gamma}{2} = 90^{\circ} - \frac{\gamma}{2}$

$\angle ZYX = \angle ZCX + \angle ZAX = \frac{\gamma}{2} + \frac{\alpha}{2} = \frac{\gamma + \alpha}{2} = \frac{180^{\circ} - \beta}{2} = 90^{\circ} - \frac{\beta}{2}$

Thus, the angles of triangle $XYZ$ are $90^{\circ} - \frac{\alpha}{2}$, $90^{\circ} - \frac{\beta}{2}$, and $90^{\circ} - \frac{\gamma}{2}$.

3. Final Answer

The angles of triangle $XYZ$ are $90^{\circ} - \frac{\alpha}{2}$, $90^{\circ} - \frac{\beta}{2}$, and $90^{\circ} - \frac{\gamma}{2}$.