Inequality Problem in a Circle

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 $$PA prove that $$\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 . Any ideas or solutions? Seemed simple at first but I've tried all the ideas I had but can't figure it out.

• 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 – Todor Markov Feb 20 at 11:09
• @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. – TlP Feb 20 at 11:17

1 Answer

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 $$\angle AQC = \angle BQC$$.

• 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? – TlP Feb 20 at 12:09
• Notice that the numerators are actually the same. – player3236 Feb 20 at 12:09
• Oh! My bad. Thanks! – TlP Feb 20 at 12:57