Question 154
In a triangle $ABC$, the internal bisector of $\angle A$ meets the side $BC$ at $D$. The lines through $D$ tangent to the circumcircles of triangles $\triangle ABD$ and $\triangle ACD$ meet the lines $AC$ and $AB$ at points $E$ and $F$, respectively. Lines $BE$ and $CF$ intersect at $G$. Prove that $\angle EDG = \angle ADF$.