The Cosine Rule for

Δ APC

and

Δ BPC

, where

∠ ACP = θ

, gives

\begin{matrix} {AP}^{2} & = {AC}^{2} + {PC}^{2} - 2 AC . PC \cos θ, \\ {PB}^{2} & = {BC}^{2} + {PC}^{2} - 2 BC . PC \cos θ . \end{matrix}

Hence

$\begin{matrix} \frac{{BC}^{2} + {PC}^{2} - {PB}^{2}}{2 BC . PC} & = \frac{{AC}^{2} + {PC}^{2} - {AP}^{2}}{2 AC . PC} = \cos θ . \end{matrix}$

Hence, multiplying both sides by $2 PC / AB$ , we find that

$\begin{matrix} \frac{{AP}^{2}}{AC . AB} + \frac{{PC}^{2}}{AB} (\frac{AC - BC}{BC . AC}) & = \frac{{PB}^{2}}{AB . BC} + \frac{AC - BC}{AB} . \end{matrix}$

As $AB + BC = AC$ , we get the result:

$\begin{matrix} \frac{{AP}^{2}}{AB . AC} + \frac{{PC}^{2}}{AC . BC} & = 1 + \frac{{PB}^{2}}{AB . BC} . \end{matrix}$