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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.

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  • $\begingroup$ This is not true. In fact $\angle APC = \angle BPC = \frac{1}{2} \angle AOC = \frac{1}{2} \angle BOC$ where $O$ is the center of the circle. See en.wikipedia.org/wiki/Inscribed_angle $\endgroup$ – Todor Markov Feb 20 at 11:09
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    $\begingroup$ @TodorMarkov you missed the part where it says $P$ is in the circle and not on it. What you are saying is true only if it's on the circle. $\endgroup$ – TlP Feb 20 at 11:17
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Extend $CP$ to $Q$.

enter image description here

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 $\angle AQC = \angle BQC$.

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  • $\begingroup$ Thanks for the quick reply and straight forward solution. How did you acquire the trigonometric inequality? I understand the equality part but how did $AP<BP$ provide us with the second inequality? $\endgroup$ – TlP Feb 20 at 12:09
  • $\begingroup$ Notice that the numerators are actually the same. $\endgroup$ – player3236 Feb 20 at 12:09
  • $\begingroup$ Oh! My bad. Thanks! $\endgroup$ – TlP Feb 20 at 12:57

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