Draw the external angle bisectors of the $A B C$ triangle. The angles of the triangle formed by these are $40^{\circ}, 65^{\circ}$ and $75^{\circ}$. What are the angles of the original triangle?
Let the vertices of the triangle formed by the external angle bisectors be denoted by \(A', B', C'\), as shown in the figure; the intersection point of the internal angle bisectors of triangle \(ABC\) is denoted by \(O\), and the angles by \(\alpha, \beta, \gamma\). We know that the internal and external angle bisectors at a vertex are perpendicular to each other, which implies that the quadrilateral \(AOBC'\) (and similarly \(AOCB'\) and \(BOCA'\)) is a cyclic quadrilateral.

In the cyclic quadrilateral \(AOBC'\), the \(65^\circ\) angle at vertex \(O\) complements the angle at \(O\) to \(180^\circ\). In the triangle \(AOB\), the sum of \(\frac{\alpha}{2}\) and \(\frac{\beta}{2}\) is therefore:

\[
\frac{\alpha}{2} + \frac{\beta}{2} = 65^\circ
\]

Similarly, from the cyclic quadrilateral \(AOCB'\):

\[
\frac{\alpha}{2} + \frac{\gamma}{2} = 75^\circ
\]

and from the cyclic quadrilateral \(BOCA'\):

\[
\frac{\gamma}{2} + \frac{\beta}{2} = 40^\circ
\]

Subtracting equation (1) from equation (2) gives \(\frac{\gamma}{2} - \frac{\beta}{2} = 10^\circ\); from equation (3), \(\frac{\gamma}{2} + \frac{\beta}{2} = 40^\circ\), hence \(\gamma = 50^\circ, \beta = 30^\circ\), and thus \(\alpha = 100^\circ\).

![](https://cdn.mathpix.com/cropped/2024_05_02_270b156ffc52a4c5ef5eg-1.jpg?height=775&width=615&top_left_y=1038&top_left_x=733)