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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ @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

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 $\angle AQC = \angle BQC$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.