Geometry In Complex Numbers The line $T$ is tangent to the circumcircle of acute triangle $ABC$ at $B.$ Let
$K$ be the projection of the orthocenter of triangle ABC onto line $T$.
Let $L$ be the midpoint of side $AC$. Show that the triangle $BKL$ is isosceles.
We are supposed to do this using complex numbers. We can easily compute $H, B, L$ But how can we compute $K$? I tried some approaches but it doesn't seem to work or I can't see how to go further. Any help will be appreciated.
 A: We may and do assume that the circumcenter $\odot (ABC)$ is the unit circle, centered in the origin $O$, and that $B$ has the affix $b=1\in \Bbb C$. Let $a,c\in\Bbb C$ be the (affix) complex representations of $A,C$, so $|a|^2=a\bar a=1$, and $|c|^2=c\bar c=1$.
We use this convention further, lower case letters are affixes for the capitalized points. So $o$, the affix of $O$ is zero. The centroid $G$ has affix $g$ with
$3g=a+b+c$.

Vectorially, $GH=-2GO$, since $G$ is placed on the segment $HO$ as it is placed on the median $BL$, so vectorially $GH+2GO=0$, passing to complex numbers
$$
(h-g)+2(o-g)=0\ ,\qquad\text{ i.e. }\qquad
h+2o=3g=a+b+c\ .
$$
In our case, $o=0$, $b=1$, so $h=a+1+c$. The line $T$ is the line parametrized by $1+it$. The point $H,K$ have the same projection on the imaginary axis, parallel to $T$, so we know $k$ if we extract and adapt the imaginary part of $h$. It is
$$
\begin{aligned}
k 
&= 1+i\operatorname{Im} h  \\
&= 1+i\operatorname{Im} (a+1+c)\\  
&= 1+i\operatorname{Im} (a+c)  \\
&= 1+2i\operatorname{Im} l  \ .
\end{aligned}
$$
Projecting on $T$, we see that $L$ projects to the mid point of $BK$, so
$\Delta LBK$ isosceles.
$\square$
