Suppose that $a, b,$ and $c$ are distinct points in $\mathbb{C}$. Let $l$ be the line which bisects the angle $\measuredangle bac$. For a point $z$ on $l$, let $p$ be the point on the line through $a$ and $b$ for which $[z, p]$ is perpendicular to $[a, b]$. For this same $z$, let $q$ be the point on the line through $a$ and $c$ for which $[z,q]$ is perpendicular to $[a, c]$.
(a) Explain why the triangles $azp$ and $azq$ are similar.
(b) Use this to show the $|z - p| = |z - q|$. This means that each point on the line bisecting $\measuredangle bac$ is equidistant to the line through $a$ and $b$ and the line through $a$ and $c.$
I am fairly certain I have part (a) using the ASA identity, however, I am not certain I have properly made the connection between part (a) and part (b).
$\textbf{Attempt:}$
Since all tree angles in a triangle must always sum to $180^{\circ}$ it is sufficient to show two of the tree angles for both our triangles have the same measure. Now, as $l$ bisects the point $a$ it follows by definition that $\measuredangle zac$ is equal to $\measuredangle zab$. Moreover, since the line $ac$ is perpendicular to the line $pz$ we know that $\measuredangle apz=90^{\circ}$. By the same reasoning we know that $\measuredangle aqz=90^{\circ}$. We have thus shown that two of the tree angles in triangle $azp$ correspond to two of the tree angles in triangle $azq$. Lastly, since both triangles share the line $az$ we know by the angle side angle identity that both triangles are similar.
Now, to see that $|z-p|=|z-q|$ let $\measuredangle paz$ be denoted by $\sin(\alpha)$ and let $\measuredangle qaz$ be denoted by $\sin(\beta)$. Then by the above argument we know that $\sin(\alpha)$ = $\sin(\beta)$. We also know that $$\sin(\alpha)=\dfrac{|z-p|}{|z-a|}$$ and $$\sin(\beta)=\dfrac{|z-q|}{|z-a|}$$ Thus,
\begin{align} \sin(\alpha) = \sin(\beta)&\Rightarrow \dfrac{|z-p|}{|z-a|}=\dfrac{|z-q|}{|z-a|}\\ &\Rightarrow |z-p|=|z-q|. \end{align}
I think it is way easier and elementary, without any trigonometry: both triangles $\;\Delta azp\;,\;\;\Delta azq\;$ are in fact congruent since both are straight triangles and besides being similar they have common hypotenuse !
So both legs are equal corr., and thus $\;|z-q|=|z-p|\;$