MathLabs

Problem 5

In a convex quadrilateral ABCDABCD, the diagonal BDBD bisects neither the angle ∠ABC\angle ABC nor the angle ∠CDA\angle CDA. The point PP lies inside ABCDABCD and satisfies ∠PBC=∠DBAand∠PDC=∠BDA.\angle PBC=\angle DBA\quad\text{and}\quad\angle PDC=\angle BDA. Prove that ABCDABCD is a cyclic quadrilateral if and only if AP=CPAP=CP.
Step 5 of 5: Finish both cases (P∉ACP \notin AC and P∈ACP \in AC)
Previous stepNext step
Case P∉AC: PA=PC  ⟺  ∡PAC=∡ACP  ⟺  ABCD cyclic;Case P∈AC: symmedian midpoint property  ⟹  (PA=PC  ⟺  ABCD cyclic)\text{Case }P\notin AC:\ PA=PC\iff\measuredangle PAC=\measuredangle ACP\iff ABCD\text{ cyclic};\quad \text{Case }P\in AC:\text{ symmedian midpoint property}\implies (PA=PC\iff ABCD\text{ cyclic})
Detailed analysis

Case one assumes P∉ACP\notin AC. Then the triangle △PAC\triangle PAC is non-degenerate, and the equal-length condition PA=PCPA=PC is the equal-base-angle condition ∡PAC=∡ACP\measuredangle PAC=\measuredangle ACP. Steps 2 and 3 reduce this to the cyclic/quasi-harmonic dichotomy; Step 4 removes the quasi-harmonic branch, so the result is cyclic ABCDABCD. Case two assumes P∈ACP\in AC. Step 4 supplies AB⋅CD=BC⋅DAAB\cdot CD=BC\cdot DA. In the cyclic direction this is DA/DC=BA/BCDA/DC=BA/BC, the symmedian criterion for BDBD in △ABC\triangle ABC; its isogonal is the median, so P is the midpoint and the equal lengths follow. Conversely the equal lengths make P the midpoint; the isogonal relation makes BD the symmedian. If D′≠BD'\ne B is the other intersection of that symmedian with the circumcircle, then D′A/D′C=BA/BCD'A/D'C=BA/BC, while the quasi-harmonic relation gives the same ratio for D. The Apollonius circle and the line through B have only the two intersections B and D', hence D=D′D=D' and cyclic ABCDABCD follows.