The angle bisectors of triangle $ABC$ intersect the circumcircle of triangle $ABC$ at points $X, Y, Z$. If $\alpha, \beta, \gamma$ are the angles of triangle $ABC$, 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

The angle bisectors of triangle $ABC$ intersect the circumcircle of triangle $ABC$ at points $X, Y, Z$. If $\alpha, \beta, \gamma$ are the angles of triangle $ABC$, 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 $O$ be the center of the circumcircle of triangle $ABC$. Let $X, Y, Z$ be the points where the angle bisectors of angles $A, B, C$ intersect the circumcircle, respectively. We are given that $\angle BAC = \alpha$, $\angle ABC = \beta$, and $\angle ACB = \gamma$. Since $AX$ is the angle bisector of angle $A$, we have $\angle BAX = \angle CAX = \frac{\alpha}{2}$.

Since angles subtended by the same arc are equal, we have $\angle BCX = \angle BAX = \frac{\alpha}{2}$ and $\angle CBX = \angle CAX = \frac{\alpha}{2}$.

The measure of arc $BX$ is equal to twice the measure of the inscribed angle $\angle BCX$. Thus, arc $BX = 2 \cdot \frac{\alpha}{2} = \alpha$. Similarly, arc $CX = 2 \cdot \frac{\alpha}{2} = \alpha$.

Since $BY$ is the angle bisector of angle $B$, we have $\angle ABY = \angle CBY = \frac{\beta}{2}$. Then arc $AY = 2 \cdot \frac{\beta}{2} = \beta$ and arc $CY = 2 \cdot \frac{\beta}{2} = \beta$.

Since $CZ$ is the angle bisector of angle $C$, we have $\angle ACZ = \angle BCZ = \frac{\gamma}{2}$. Then arc $AZ = 2 \cdot \frac{\gamma}{2} = \gamma$ and arc $BZ = 2 \cdot \frac{\gamma}{2} = \gamma$.

Now consider the arcs that form the circumcircle: $BX = CX = \alpha$, $AY = CY = \beta$, and $AZ = BZ = \gamma$.

Arc $YZ = arc YC + arc CZ = \beta + \gamma$.

Then $\angle YXZ = \frac{1}{2} arc YZ = \frac{1}{2} (\beta + \gamma)$. Since $\alpha + \beta + \gamma = 180^{\circ}$, we have $\beta + \gamma = 180^{\circ} - \alpha$. Thus, $\angle YXZ = \frac{1}{2} (180^{\circ} - \alpha) = 90^{\circ} - \frac{\alpha}{2}$.

Similarly, arc $XZ = arc XC + arc CZ = \alpha + \gamma$.

Then $\angle XYZ = \frac{1}{2} arc XZ = \frac{1}{2} (\alpha + \gamma)$. Since $\alpha + \beta + \gamma = 180^{\circ}$, we have $\alpha + \gamma = 180^{\circ} - \beta$. Thus, $\angle XYZ = \frac{1}{2} (180^{\circ} - \beta) = 90^{\circ} - \frac{\beta}{2}$.

Finally, arc $XY = arc XA + arc AY = arc XB + arc AY = \alpha + \beta$.

Then $\angle XZY = \frac{1}{2} arc XY = \frac{1}{2} (\alpha + \beta)$. Since $\alpha + \beta + \gamma = 180^{\circ}$, we have $\alpha + \beta = 180^{\circ} - \gamma$. Thus, $\angle XZY = \frac{1}{2} (180^{\circ} - \gamma) = 90^{\circ} - \frac{\gamma}{2}$.

3. Final Answer

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