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 $\overleftrightarrow{BH}$ is fixed, regardless of the location of point $P$ on $l$. In other words, the intersection Q of line $\overleftrightarrow{BH}$ with the line $l$ is a fixed point. Image for reference:

[![enter image description here](https://i.stack.imgur.com/e9uDw.png)](https://i.stack.imgur.com/e9uDw.png)

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.