Let $\overline{OH}$ cut $\overline{BC}, \overline{CA}$ and $\overline{AB }$ at $X, Y$ and $Z $. From a simple angle chase, we have that $l_a $ splits $\measuredangle BAC $ the same way $\overline{OH} $ splits $\measuredangle BHC $. Now just use the ratio lemma on $\triangle BHC, \triangle CHA$ and $\triangle AHB$ and Menelaus on $\triangle ABC $ with $\overline{OHXYZ} $ to get: $\prod_{cyc} \frac{sin \measuredangle (\overline{AB}, l_a)}{sin \measuredangle (\overline{AC}, l_a)} = \prod_{cyc} \frac{sin \measuredangle (\overline{BH}, \overline{OH})}{sin \measuredangle (\overline{CH}, \overline{OH})} = \prod_{cyc} \frac{|BH|}{|CH|} \cdot \frac{|BX|}{|CX|} = 1$. Thus by Trigonometric Ceva, we are done.

Interesting problem. Here is my solution First, notice that $H$ is the centre of $\odot (WUOV)$ and $UO || AC$, $WO || BC$, $OV || AB$. Easy angle chase gives us that $\triangle WUV \sim \triangle ABC$. Let $D$ be a point on $\odot (ABC)$, such that $WUOV \sim ABDC$ Now, I will prove that $BD || WV$. $\angle CBD = \angle VUO = \angle OWV$. Because $WO || BC$, we get that $BD || WV$. Similar, $AD || UV$ and $CD || WU$. Thus, $\ell_A$, $\ell_B$, $\ell_C$ are concurrent at $D$ $\blacksquare$

Attachments:

@above wonderful solution! Here's mine my complex numbers: First, we define $H_A$ as the second intersection of the $A$ altitude with circumcircle. We know that $h_a = -\bar abc$. Therefore, we know that $w = a-\bar abc$, $u = b-a\bar bc$, and $v = c-ab\bar c$. Then, we claim that the concurrent point lies on the unit circle, so we find the second intersection of the point of the line through $A$ parallel to $UV$ with the unit circle. Let $X$ be this point, we compute that \[\frac{x-a}{u-v} \in \mathbb R \implies \frac{x-a}{u-v} = \frac{a-x}{ax(\bar u - \bar v)} \implies ax(\bar v -\bar u) = u-v \implies x = \frac{u-v}{a(\bar v - \bar u)} \]In particular, we simplify that \begin{align*} x &= \frac{u-v}{a(\bar v - \bar u)} \\ &= \frac{b-a\bar b c - c + ab \bar c}{a(\bar c - \bar a \bar b c - \bar b + \bar a b \bar c)} \\ &= \frac{a(b^2-c^2)+bc(b-c)}{b^2-c^2+a(b-c)} \\ &= \frac{ab+ac+bc}{a+b+c} \end{align*}Note that this expression is symmetrical in terms of $a,b,c$. Therefore, we see that this is the desired point of the intersection.