Let ω be a circle with centre O. Let γ be another circle passing through O and intersecting ω at points A and B. A diameter CD of ω intersects γ at a point P different from O. Prove that ∠APC=∠BPD
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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 γ). Best regards, sunken rock
20.11.2015 22:26
Suppose that A0 is a point on γ such that ∠A0PC = ∠BPD. Then the segments OA0 and OB subtends same angle in the respective minor arcs, so OA0 = OB. This shows that A lies on ω and hence A0=A. This proves that ∠APC= ∠BPD.
04.12.2015 12:07
My Solution. Clearly, OPAB is cyclic, as O,P,A,B∈γ. Thus, ∠ABO=180∘−∠OPA=∠APC. And, ∠BPD=∠OPB=∠OAB. As A,B∈ω, and O is the center of ω, we can say that OA=OB, which implies that ∠OAB=∠OBA. Now, we can say that ∠BPD=∠OAB=∠ABO=∠APC⇒∠APC=∠BPD . \Square
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04.12.2015 13:21
Inverting the problem through ω directly solves it
10.02.2019 10:37
Let ∠OBA=∠OAB=∠OPB=x ⟹ ∠BOA=∠BPA=180∘−2x ⟹ ∠OPB=∠DPB=x=∠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.