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'Cosines Rule' printed from https://nrich.maths.org/
The Cosine Rule for $\Delta APC$ and $\Delta BPC$, where $\angle
ACP=\theta$, gives $\begin{eqnarray} AP^2 &= AC^2+PC^2-2AC.PC
\cos\theta,\cr PB^2&= BC^2+PC^2-2BC.PC \cos \theta.
\end{eqnarray}$
Hence $\begin{eqnarray} \frac{BC^2+PC^2-PB^2}{ 2BC.PC}&=
\frac{AC^2+PC^2-AP^2}{ 2AC.PC} = \cos\theta. \end{eqnarray}$
Hence, multiplying both sides by $2PC/AB$, we find that
$\begin{eqnarray} {AP^2\over AC.AB} +{PC^2\over
AB}\left({AC-BC\over BC.AC}\right) &= {PB^2\over AB.BC}
+{AC-BC\over AB}.\end{eqnarray}$ As $AB+BC=AC$, we get the result:
$\begin{eqnarray} {AP^2\over AB.AC}+{PC^2\over AC.BC} &= 1 +
{PB^2\over AB.BC}. \end{eqnarray}$