Suppose that $A, B$, and $C$ are distinct points in $C$. Suppose that $l$ is the line which bisects $∠BAC$ and $M$ is the line which bisects $∠ACB$ Suppose that $A, B$, and $C$ are distinct points in $C$. Suppose that $l$ is the line which bisects $∠BAC$ and $M$ is the line which bisects $∠ACB$. Now let $Z$ be the point $l \cap m$. Let $n$ be the line through $B$ and $Z$. Show that $N$ bisects $∠CBA$. This shows that the angle bisectors of a triangle are concurrent.

$\textbf{Attempt:}$
Construct perpendicular lines from the point $Z$ to the lines $AB,BC$ and $AC$ and label the new point $E,F$ and $G$ respectively.

Now, angle $GZC$ is congruent to angle $FZC$ by the angle side angle identity. This implies that $ZG=ZF\quad (\star)$.
Also, angle $EZA$ is congruent to angle FZA by the angle side angle identity which implies $ZE=ZF\quad(\star\star).$
Putting $(\star)$ together with $(\star\star)$ shows that $ZG=ZE=ZF.$
Now construct a line from $B$ to $Z$ and label this new line $n$

By our construction angles $BGZ$ and $BEZ$ are both right angles. Moreover, we may conclude they are congruent angles as the hypotenuse and adjacent sides are equal. Thus, $BGZ=BEZ$ which shows that $n$ is the angle bisector.
I am not entirely sure I have answered this question properly, however, I do not feel there are any flaws in my logic. Would someone please look over and make sure I am on the correct path.
Also, we have not discussed the angle side angle (ASA) identity in class. I did however pick this up from the text we are reading from. wold it be fair game to use this identity or is there another way to go about this problem. Thank you.
 A: Your writing an angle can be confused. When you refer to an angle of $\measuredangle PQR$ I understand that you are referring to  angle determined by the intersection of $PR$ and $QR$ segments.
Your attempt begins properly:

Construct perpendicular lines from the point $z$ to the lines $AB$, $BC$,  and $AC$ and label the new point $E$, $F$, and $G$ respectively.
Now, angle $\measuredangle GZC$ is congruent to angle $\measuredangle FZC$ by the angle side snuggle identity. This implies that $ZG=ZF$ $(⋆)$.

Actually $\measuredangle GZC$ is congruent to angle $\measuredangle FZC$ by Angle Angle Side postulate, also,  $ZG=ZF$  is due to Angle Side Angle postulate.

Also, angle $\measuredangle EZA$ is congruent to angle $\measuredangle FZA$ by the angle side angle identity which implies $ZE=ZF$ (⋆⋆).

In the same way, $\measuredangle EZA$ is congruent to angle $\measuredangle FZA$ by Angle Angle Side postulate, also,  $ZE=ZF$ is due to Angle Side Angle postulate.

Putting (⋆) together with (⋆⋆) shows that $ZG=ZE=ZF$.

The latter is fine.

By our construction angles $\measuredangle BGZ$ and $\measuredangle BEZ$ are both right angles. Moreover, we may conclude they are congruent angles as the hypotenuse and adjacent sides are equal. Thus, $\measuredangle BGZ=\measuredangle BEZ$ which shows that $n$ is the angle bisector.

The conclusion you get is due to you have two right triangles with the same hypotenuse and two equal sides, then by the Pythagoras Theorem  the other pair of sides are also  equals, i.e, $EB=BF$. Therefore,  $\measuredangle BGZ=\measuredangle BEZ$ by Side Side Side postulate.
Overall your argument is correct.
