Let $\omega$ be the circumcircle of triangle $ABC$. A point $B_1$ is chosen on the prolongation of side $AB$ beyond point B so that $AB_1 = AC$. The angle bisector of $\angle BAC$ meets $\omega$ again at point $W$. Prove that the orthocenter of triangle $AWB_1$ lies on $\omega$ . (A.Tumanyan)
Problem
Source: 2012 Sharygin Geometry Olympiad Final Round 8.6
Tags: geometry, circumcircle, orthocenter