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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.

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    $\begingroup$ When you say "$\angle ABG$ and $\angle FBG$ are the same", do you assume that $G$ is on the bisector of $\angle B$? But this is what you are to prove. $\endgroup$ – A.Γ. Jun 17 '18 at 12:35
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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.

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My attempt -

We extend $BE$ to $AC$ with intersection at $G'$. We then construct the perpendicular $G'F'$ from $G'$ to $BC$. If we can show that $AF'$ bisects $\angle DAC$, we are done. (Why?)

We see that $\angle BG'A=90^\circ - \angle G'BA = \dfrac {\angle B}2$ and $\angle BG'F'=\angle G'BF' = \dfrac {\angle B}2$.

To prove that $AF'$ indeed bisects $\angle DAC$, I leave as an exercise.

Should you need a hint -$\\$

$\\$

Find $\angle DAC$ and $\angle DAF'$ in terms of $\angle B$

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