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.
The bisector of $\angle B$ meets the bisector of $\angle CAD$ at the point, call it $H$.
$\triangle DAC\simeq\triangle ABC$ by AA similarity $\Rightarrow$ $\angle B=\angle CAD$ and $\angle FAD=\angle EBD$. Then $\triangle BDE\simeq\triangle AHE$ by AA similarity. Hence, the bisectors are orthogonal and $H$ is the midpoint of $AF$.
$AD\|GF$, hence, $\angle DAF=\angle AFG$. Therefore, $\angle FAG=\angle AFG$ and $\triangle GAF$ is isosceles. Thus, the segment from $G$ to the midpoint $H$ is the altitude, and $BH$ and $GH$ are on the same line.