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 1 of 5: Isogonal conjugation in △PBD\triangle PBD
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P∉BD,∠PBC=∠DBA, ∠PDC=∠BDA  ⟹  A,C isogonal conjugates in △PBD  ⟹  ∠APB+∠CPD=180∘P\notin BD,\quad \angle PBC=\angle DBA,\ \angle PDC=\angle BDA \implies A,C\text{ isogonal conjugates in }\triangle PBD \implies \angle APB+\angle CPD=180^\circ
Detailed analysis

The non-bisecting hypotheses imply P∉BDP\notin BD: if P were on BD, the first given angle would read ∠PBC=∠DBC\angle PBC=\angle DBC, making BD bisect the angle at B. Thus △PBD\triangle PBD is non-degenerate. The first condition makes BA and BC isogonal at B, and the second makes DA and DC isogonal at D. The elementary three-cevian isogonality theorem follows by reflecting the first two pairs in their angle bisectors; it makes PA,PCPA,PC isogonal at P. Since P is interior, ∠APB+∠CPD=180∘\angle APB+\angle CPD=180^\circ.