Proof on an equilateral triangle with a cevian extended to its circumcircle

Proof on an equilateral triangle with a cevian extended to its circumcircle

Consider the following figure with equilateral triangle $ABC$ and a cevian
$AQ$ extended to $P$ on its circumcircle.

We are required to prove that:
$\frac{1}{PB} + \frac{1}{PC} = \frac{1}{PQ}$
Let $\angle PAC = \alpha$ and let length of $AB = s$
By the Law of Sines,
$\frac{PC}{\sin\alpha} = \frac{s}{\sin\angle CPA} =
\frac{s}{\sin60^{\circ}} \implies PC = \frac{s\sin\alpha}{\sin60^{\circ}}$
Similarly, $PB = \frac{s\sin(60^{\circ} - \alpha)}{\sin60^{\circ}}$
$\frac{1}{PB} + \frac{1}{PC} = \frac{\sin60}{s}\left(\frac{\sin(60 -
\alpha) + \sin\alpha}{\sin\alpha\sin(60 - \alpha)}\right)$
It remains to be proven that:
$PQ = \frac{\sin60\sin(60 - \alpha)\sin\alpha}{\sin60\sin(60 - \alpha) +
\sin60\sin\alpha}$
I'm utterly lost from here.