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 2 of 5: Rewrite ∠APB+∠CPD=180∘\angle APB + \angle CPD = 180^\circ in terms of △PAC\triangle PAC
Previous stepNext step
∡APB+∡CPD=0  ⟺  ∡PAC−∡ACP=∡BAC+∡CBD+∡DCA+∡ADB\measuredangle APB+\measuredangle CPD=0 \iff \measuredangle PAC-\measuredangle ACP=\measuredangle BAC+\measuredangle CBD+\measuredangle DCA+\measuredangle ADB
Detailed analysis

Use directed angles modulo 180∘180^\circ. Step 1 gives ∡APB+∡CPD=0\measuredangle APB+\measuredangle CPD=0. Completing the two given angles at B and D gives ∡PBA=∡CBD\measuredangle PBA=\measuredangle CBD and ∡PDC=∡ADB\measuredangle PDC=\measuredangle ADB. Decomposing the two angles at P in triangles ABP and CDP gives ∡APB=∡PBA+∡BAP=∡CBD+∡BAC−∡PAC\measuredangle APB=\measuredangle PBA+\measuredangle BAP=\measuredangle CBD+\measuredangle BAC-\measuredangle PAC and ∡CPD=∡PDC+∡DCP=∡ADB+∡DCA+∡ACP\measuredangle CPD=\measuredangle PDC+\measuredangle DCP=\measuredangle ADB+\measuredangle DCA+\measuredangle ACP. Adding and using Step 1 yields ∡PAC−∡ACP=∡BAC+∡CBD+∡DCA+∡ADB\measuredangle PAC-\measuredangle ACP=\measuredangle BAC+\measuredangle CBD+\measuredangle DCA+\measuredangle ADB, hence ∡PAC=∡ACP\measuredangle PAC=\measuredangle ACP exactly when the four-angle sum is zero.