Let $\omega$ be a circle with centre $O$. Let $\gamma$ be another circle passing through $O$ and intersecting $\omega$ at points $A$ and $B$. $A$ diameter $CD$ of $\omega$ intersects $\gamma$ at a point $P$ different from $O$. Prove that $\angle APC= \angle BPD$
Problem
Source:
Tags: geometry
13.12.2013 13:14
13.12.2013 19:49
$OA=OB$ is enough to solve the problem (they are both chords of $\gamma$). Best regards, sunken rock
20.11.2015 22:26
Suppose that $A_0$ is a point on $\gamma$ such that ∠$A_0PC$ = ∠$BPD$. Then the segments $OA_0$ and $OB$ subtends same angle in the respective minor arcs, so $OA_0$ = $OB$. This shows that A lies on $\omega$ and hence $A_0 = A$. This proves that ∠$APC$= ∠$BPD$.
04.12.2015 12:07
My Solution. Clearly, $OPAB$ is cyclic, as $O,P,A,B\in \gamma.$ Thus, $\angle ABO = 180^{\circ}-\angle OPA=\angle APC.$ And, $\angle BPD=\angle OPB=\angle OAB.$ As $A,B\in \omega,$ and $O$ is the center of $\omega,$ we can say that $OA=OB,$ which implies that $\angle OAB=\angle OBA.$ Now, we can say that $\angle BPD=\angle OAB=\angle ABO=\angle APC\Rightarrow \boxed{\angle APC=\angle BPD}~.~\Square$
Attachments:

04.12.2015 13:21
Inverting the problem through $\omega $ directly solves it
10.02.2019 10:37
Let $\angle OBA=\angle OAB=\angle OPB=x $ $ \implies $ $\angle BOA=\angle BPA=180^{\circ}-2x $ $\implies $ $\angle OPB=\angle DPB=x=\angle APC $
10.02.2019 11:20
cute(ugly) solution : invert the diagram with P and denote A becoming A (for we don't need old points) . since the inversion don't change ∠(omega,line CD) ,CD is still the diameter of the reflection of omega . since DCPO are harmonic and ∠DBC is 90°(∵diameter CD), ∠ABD=∠PBD . On the other hand ,∠ABD=∠BCD, so ∠PBD=∠BCD ,showing the intersection of BD and AC point G and P,C,B lie on the same circle . In the same way we know G,P,A,D lie on the same circle , so ∠APD=∠AGD=∠BGC=∠BPC ,which is equivalent to the original problem.