3. The angle bisectors of angles A and C divide the non-isosceles triangle ABC into a quadrilateral and three triangles, and among these three triangles, there are two isosceles triangles. Find the angles of triangle $A B C$.
Answer. $180^{\circ} / 7,2 \cdot 180^{\circ} / 7,4 \cdot 180^{\circ} / 7$. Solution. Let $AK$ and $CM$ be the angle bisectors, and $I$ their point of intersection. Let's see which angles in which triangles can be equal. Triangle $AIC$ cannot be isosceles: in it, $\angle AIC = 90^{\circ} + \angle ABC / 2$ is obtuse, while $\angle IAC = \angle BAC / 2 \neq \angle BCA / 2 = \angle ICA$. It is also impossible for $\angle MIA = \angle MAI$ and $\angle KCI = \angle KIC$, because the external angle of triangle $AIC$ cannot be equal to an internal angle not adjacent to it. Finally, it is impossible for both $\angle MIA = \angle IMA$ and $\angle KIC = \angle IKC$ to hold simultaneously, or for both $\angle IAM = \angle IMA$ and $\angle ICK = \angle IKC$ to hold simultaneously, because then $\angle BAC = \angle ACB$. Therefore, the only possibility (up to the permutation of points $A$ and $C$) is:

$$
\angle AIM = \angle AMI = 90^{\circ} - \angle ABC / 2 = 90^{\circ} - \angle BAC / 4, \angle ICK = \angle IKC = \angle ACB / 2 = 90^{\circ} + \angle ABC / 2 - \angle ACB / 2.
$$

This means that $2 \angle ABC = \angle BAC$ and $\angle ACB = 90^{\circ} + \angle ABC / 2$. Then $7 \angle ABC / 2 = 90^{\circ}$, which leads to the answer.