<s>[INST] 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. 
 [/INST] by Ptolemy ,  $AB\cdot PC+AC\cdot PB=BC\cdot AP$
$\therefore PA=PB+PC$
$\triangle PCQ\sim\triangle PAB$  $\Longrightarrow$ $PQ\cdot PA=PB\cdot PC$
$PQ\cdot PA=PQ\cdot (PB+PC)=PB\cdot PC$
$\therefore$ $\dfrac{1}{PB}+\dfrac{1}{PC}=\dfrac{1}{PQ}$
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