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. 

I have read that this problem appeared in South African mathematical competition in years 1994 and 1997. The proof I know is based on the areas of triangles. 
As $\angle APB=\angle APC=60^\circ$ and $\angle BPC=120^\circ$, the areas of triangles $BPQ,$ $QPC$ and $BPC$ are $\frac{1}{2}BP\cdot PQ\cdot \sin 60^\circ,$ $\frac{1}{2}PQ\cdot PC\cdot \sin 60^\circ$
and $\frac{1}{2}BP\cdot PC\cdot \sin 120^\circ$ respectively. As $\sin 60^\circ=\sin 120^\circ$ and the sum of areas of triangles $BPQ$ and $QPC$ is the area
of $BPC,$ we have that $BP\cdot PQ+PQ\cdot PC=BP\cdot PC.$ Now divide this equation by $BP\cdot PQ\cdot PC.$
