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Problem 6

A convex quadrilateral ABCDABCD satisfies AB⋅CD=BC⋅DAAB\cdot CD=BC\cdot DA. Point XX lies inside ABCDABCD so that ∠XAB=∠XCD\angle XAB=\angle XCD and ∠XBC=∠XDA\angle XBC=\angle XDA. Prove that ∠BXA+∠DXC=180∘\angle BXA+\angle DXC=180^\circ.
Step 3 of 8: Invert the quadrilateral at XX
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In plain words

Inversion changes each distance by endpoint factors, which cancel in the product condition.

P′Q′=r2 PQXP⋅XQP'Q'=\frac{r^2\,PQ}{XP\cdot XQ}
Detailed analysis

Let A′,B′,C′,D′A',B',C',D' be the inverse images of A,B,C,DA,B,C,D under an inversion of radius rr centered at XX. The distance formula is P′Q′=r2 PQXP⋅XQP'Q'=\frac{r^2\,PQ}{XP\cdot XQ}. Hence A′B′⋅C′D′=B′C′⋅D′A′A'B'\cdot C'D'=B'C'\cdot D'A' because both sides equal r4AB⋅CDXA XB XC XD\frac{r^4 AB\cdot CD}{XA\,XB\,XC\,XD}. The inverted quadrilateral is again quasi-harmonic.