amparvardi wrote: Points $A', B', C'$ lie on sides $BC, CA, AB$ of triangle $ABC.$ for a point $X$ one has $\angle AXB =\angle A'B'C' + \angle ACB$ and $\angle BXC = \angle B'A'C' +\angle BAC.$ Prove that the quadrilateral $XA'BC'$ is cyclic. I think $\angle AXB=\angle A'C'B'+\angle ACB$. Let $P$ be the intersection of the circumcircles of triangles $AB'C',A'BC',A'B'C$. Then $\angle APB=\angle ACB+\angle PAC+\angle PBC=\angle ACB+\angle PC'B'+\angle PC'A'=\angle ACB+\angle A'C'B'$. Similarly $\angle BPC=\angle BAC+\angle B'A'C'$. Thus $P\equiv X$ and we are done.

oneplusone wrote: amparvardi wrote: Points $A', B', C'$ lie on sides $BC, CA, AB$ of triangle $ABC.$ for a point $X$ one has $\angle AXB =\angle A'B'C' + \angle ACB$ and $\angle BXC = \angle B'A'C' +\angle BAC.$ Prove that the quadrilateral $XA'BC'$ is cyclic. I think $\angle AXB=\angle A'C'B'+\angle ACB$. Oh, sorry. You are right. I corrected the problem. Thank you.