In the triangle $ABC$, $\angle A = 90$°, the bisector of $\angle B$ meets the altitude $AD$ at the point $E$ and the bisector of $\angle CAD$ meets the side $CD$ at $F$. The line through $F$ perpendicular to $BC$ intersects $AC$ at $G$. Prove that $B,E,G$ are collinear.
So I did write a proof (below) but I was wondering if I did it right — if I got it wrong and there's another way to prove the collinearity, or if I need to add on/take out something etc., please point that out.
$\triangle BAG \equiv \triangle BFG$ by AAS similarity: $\angle ABG$ and $\angle FBG$ are the same, $\angle BAG$ and $\angle BFG$ are both $90$° (these two facts are basically given in the problem) and the triangles share a side $BG$.
Since the points $B$ and $G$ are endpoints of a side of both triangles — $BG$ — they must be collinear.
Then we note that $AD//GF$ because $\angle BDA = \angle BFG = 90$°. Therefore, $\angle BED = \angle BGF$ also. So by AA similarity, $\triangle DBE \simeq \triangle FBG$. Since the triangles share angle $\angle DBE$ and is thus nested within each other, $BE$ must be on $BG$, meaning point $E$ must be on line $BG$. Therefore, $B,E,G$ are collinear.