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 4 of 5: Quasi-harmonic iff PP lies on ACAC
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X:=AC∩BD,Y,Y′:=isogonals of BD in ∠B,∠D on AC  ⟹  AX⋅AYCX⋅CY=BA2BC2, AX⋅AY′CX⋅CY′=DA2DC2  ⟹  (AB⋅CD=BC⋅DA  ⟺  P∈AC)X:=AC\cap BD,\quad Y,Y':=\text{isogonals of }BD\text{ in }\angle B,\angle D\text{ on }AC \implies \frac{AX\cdot AY}{CX\cdot CY}=\frac{BA^2}{BC^2},\ \frac{AX\cdot AY'}{CX\cdot CY'}=\frac{DA^2}{DC^2} \implies (AB\cdot CD=BC\cdot DA \iff P\in AC)
Detailed analysis

Let X=AC∩BDX=AC\cap BD and let the isogonal ray from B meet AC at Y. The sine rule gives AXCX=ABsin⁡∠ABXBCsin⁡∠XBC\frac{AX}{CX}=\frac{AB\sin\angle ABX}{BC\sin\angle XBC} and, after swapping the two sines for the isogonal ray, AYCY=ABsin⁡∠XBCBCsin⁡∠ABX\frac{AY}{CY}=\frac{AB\sin\angle XBC}{BC\sin\angle ABX}. Multiplication gives AX⋅AYCX⋅CY=BA2BC2\frac{AX\cdot AY}{CX\cdot CY}=\frac{BA^2}{BC^2}. Let the corresponding isogonal ray from D meet AC at Y'. The same calculation gives AX⋅AY′CX⋅CY′=DA2DC2\frac{AX\cdot AY'}{CX\cdot CY'}=\frac{DA^2}{DC^2}. The directed ratio AT/CTAT/CT determines a point T on AC, so Y and Y' coincide exactly when the right sides agree, namely AB⋅CD=BC⋅DAAB\cdot CD=BC\cdot DA. Since P is the intersection of the two isogonal rays, this is equivalent to P∈ACP\in AC.