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. 

Observe that the cuadrilateral $ABPC$ is cyclic, so by similarity we have $\triangle PQB \sim \triangle CQA$ y $\triangle PQC \sim \triangle BQA$. Therefore:
\begin{equation*}
  \dfrac{PQ}{PB} = \dfrac{QC}{AC} \qquad \text{y} \qquad \dfrac{PQ}{PC} = \dfrac{BQ}{AB}.
\end{equation*}
Thus, taking into account that $AB = AC = BC$, we have:
\begin{equation*}
  \dfrac{PQ}{PB} + \dfrac{PQ}{PC} = \dfrac{QC}{AC} + \dfrac{BQ}{AB} = \dfrac{QC + BQ}{AB} = \dfrac{BC}{AB} = 1. 
\end{equation*}
Dividing by $PQ$, we conclude:
\begin{equation*}
 \dfrac{1}{PB} + \dfrac{1}{PC} = \dfrac{1}{PQ}.
\end{equation*}
