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 3 of 5: Interpret the four-angle condition via complex cross-ratios
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∡BAC+∡CBD+∡DCA+∡ADB=0  ⟺  z+1z∈R for z=(a−b)(c−d)(b−c)(a−d)  ⟺  ABCD cyclic or AB⋅CD=BC⋅DA\measuredangle BAC+\measuredangle CBD+\measuredangle DCA+\measuredangle ADB=0 \iff z+\frac{1}{z}\in\mathbb{R}\ \text{for }z=\frac{(a-b)(c-d)}{(b-c)(a-d)} \iff ABCD\text{ cyclic or }AB\cdot CD=BC\cdot DA
Detailed analysis

Put the vertices at complex coordinates a,b,c,da,b,c,d and define z=(a−b)(c−d)(b−c)(a−d)z=\frac{(a-b)(c-d)}{(b-c)(a-d)}. Direct expansion gives (a−b)(c−d)+(b−c)(a−d)=(a−c)(b−d)(a-b)(c-d)+(b-c)(a-d)=(a-c)(b-d), so z+1=(a−c)(b−d)(b−c)(a−d)z+1=\frac{(a-c)(b-d)}{(b-c)(a-d)}. The product of the four complex quotients whose arguments are ∡BAC,∡CBD,∡DCA,∡ADB\measuredangle BAC,\measuredangle CBD,\measuredangle DCA,\measuredangle ADB is, up to a real sign, (z+1)2/z=z+2+1/z(z+1)^2/z=z+2+1/z. Hence the four-angle sum is 00 modulo 180∘180^\circ exactly when z+1/zz+1/z is real. Writing z=x+iyz=x+iy, its imaginary part is y(1−1/∣z∣2)y(1-1/|z|^2); therefore this happens exactly when z∈Rz\in\mathbb R or ∣z∣=1|z|=1. The first condition is the usual cross-ratio criterion for A,B,C,DA,B,C,D to be cyclic. The second is ∣a−b∣∣c−d∣=∣b−c∣∣a−d∣|a-b||c-d|=|b-c||a-d|, namely AB⋅CD=BC⋅DAAB\cdot CD=BC\cdot DA, the quasi-harmonic condition.