$G_1$ is the axisymmetric image of focus $F$ with respect to one tangent, and $G_2$ with respect to the other one. Let these tangents intersect at point $B$ on the directrix of the given parabola.
Because $\angle D_2BG_2=\angle GBG_1$ and $\angle D_2BF = \angle D_2BG_2$ , it follows that $\angle GBG_1= \angle D_2BF$. Since $\angle BG_1G=\angle BFD_2=90°$, and $|BF|=|BG_1|$ , $\triangle FD_2B$ is congruent to $\triangle G_1GB$, which implies that $|D_2B|=|BG|$ and $\angle FD_2B=\angle G_1GB$. So now we know that the midpoint of $D_2G$ is $B$ and $\triangle D_2GD_1$ is isosceles which means that line $D_1B$ is a perpendicular bisector of $D_2G$.
Is this proof okay and are there any more elegant proofs?