so here is a question I came across scanning thru geometry questions:
$A$ and $B$ are points on a circle and $C$ is the midpoint of arc $\overset{\LARGE \frown}{AB}$. P is an arbitrary point IN the circle where $ PA<PB$ prove that $\angle BPC< \angle APC$.
I tried extending $BP$ and $AP$ and calculating the angles using arc relations but I still get stuck and can't find a good way to use the fact that $ PA<PB$ . Any ideas or solutions? Seemed simple at first but I've tried all the ideas I had but can't figure it out.
Extend $CP$ to $Q$.
We have $\angle AQC = \angle BQC$ and $AP < BP$.
We also have $\dfrac {\sin\angle PAQ}{PQ} = \dfrac {\sin \angle AQP}{AP} > \dfrac {\sin \angle BQP}{BP} = \dfrac {\sin\angle PBQ}{PQ}$.
Additionally, note that $\angle CBQ$, and hence $\angle PBQ$, is always acute.
Hence $\angle QAP > \angle QBP$, and thus $\angle APC > \angle BPC$ by considering exterior angles and $\angle AQC = \angle BQC$.