Let $D, E, F$ be the points where the inscribed circle touches the sides of triangle $ABC$. Prove that the angles of triangle $DEF$ are $90^{\circ} - \frac{\alpha}{2}$, $90^{\circ} - \frac{\beta}{2}$, and $90^{\circ} - \frac{\gamma}{2}$, where $\alpha, \beta, \gamma$ are the angles of triangle $ABC$.

1. Problem Description

Let $D, E, F$ be the points where the inscribed circle touches the sides of triangle $ABC$.

Prove that the angles of triangle $DEF$ are $90^{\circ} - \frac{\alpha}{2}$, $90^{\circ} - \frac{\beta}{2}$, and $90^{\circ} - \frac{\gamma}{2}$, where $\alpha, \beta, \gamma$ are the angles of triangle $ABC$.

2. Solution Steps

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

Since $D, E, F$ are points of tangency, $ID \perp BC$, $IE \perp AC$, and $IF \perp AB$.

Consider quadrilateral $AFIE$. We have $\angle AFI = \angle AEI = 90^{\circ}$.

The sum of angles in a quadrilateral is $360^{\circ}$, so

$\angle FIE = 360^{\circ} - \angle AFI - \angle AEI - \angle FAE = 360^{\circ} - 90^{\circ} - 90^{\circ} - \alpha = 180^{\circ} - \alpha$.

Since $\angle FIE$ is the central angle subtending the arc $FE$ in the circumcircle of triangle $DEF$, and $\angle FDE$ is an inscribed angle subtending the same arc, we have $\angle FDE = \frac{1}{2} \angle FIE = \frac{1}{2} (180^{\circ} - \alpha) = 90^{\circ} - \frac{\alpha}{2}$.

Similarly, consider quadrilateral $BDIF$. We have $\angle BDI = \angle BFI = 90^{\circ}$.

So $\angle DIB = 360^{\circ} - 90^{\circ} - 90^{\circ} - \beta = 180^{\circ} - \beta$.

Thus $\angle DEF = \frac{1}{2} \angle DIB = \frac{1}{2}(180^{\circ} - \beta) = 90^{\circ} - \frac{\beta}{2}$.

Finally, consider quadrilateral $CEID$. We have $\angle CEI = \angle CDI = 90^{\circ}$.

So $\angle EID = 360^{\circ} - 90^{\circ} - 90^{\circ} - \gamma = 180^{\circ} - \gamma$.

Thus $\angle EFD = \frac{1}{2} \angle EID = \frac{1}{2}(180^{\circ} - \gamma) = 90^{\circ} - \frac{\gamma}{2}$.

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

3. Final Answer

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