Question 67
Let $k$ be the circumcircle of $\triangle ABC$ with $\angle ACB \gt 90^\circ$, and $BD$ be the diameter of $k$ through $B$. The circle $k_1$ with center $D$ and radius $DC$ meets $k$ at point $E$ and $AB$ at point $G$. If $F$ is the intersection point of $GE$ and $BD$, prove that $\angle DCG = \angle EFD$.