Vtyjkyuk

Let $ABC$ be a given equilateral triangle. From vertex $C$ trace a line $l$ with a fixed angle $alpha$. Now, let $P$ be an arbitrary point on $l$ and construct the equilateral triangle with side $AP$ which has $H$ as its third vertex.

What I want to prove (or know why does it happen) is that the line $overleftrightarrowBH$ is fixed, regardless of the location of point $P$ on $l$. In other words, the intersection Q of line $overleftrightarrowBH$ with the line $l$ is a fixed point. Image for reference:

Another way of seeing this problem is that point $H$ describes a straight line while $P$ moves through $l$.
PS:It seems that angle $BQP$ is of $60°$. I'm lost about what path should I follow for the proof.

Let's look at the geometric transformation in the plane given by the clockwise rotation of 60 degrees in the plan around the point $A$. This transformation is an isometry mapping lines to lines. In particular, it maps the line $l$ to another line $m$ that makes the same angle of 60 degrees with $l$. Since the transformation maps $C$ to $B$ and $C$ is on $l$, so the line $m$ must pass through $B$.

I try to explain it from another angle.

1) Form the circum-circle of ABC. Let it cut “L” at Q such that ABQC is cyclic with $angle AQB = angle ACB = 60^0$.

2) Let P be any point on “L”. Join AP. Locate the point H on QB extended such that $angle APH = 60^0$.

3) After joining AH, we found that AQPH is also cyclic.

Note that $angle HAP = angle H(B)QP = angle BAC = 60^0$. This means the triangle APH thus created is equilateral.