MathLabs

Problem 4

Let ABCDABCD be a quadrilateral with an incircle ω\omega of centre II. The diagonals ACAC and BDBD intersect at EE. Let JJ be the incentre of triangle ABDABD. The extension of the ray EJEJ intersects ω\omega at PP. Prove that PI⊥BDPI\perp BD.
Step 4 of 6: A harmonic range from the pole–polar configuration
V:=IL∩BD  ⟹  (P′,V,Q,L) harmonicV:=IL\cap BD \implies (P',V,Q,L)\text{ harmonic}
Detailed analysis

Let V:=IL∩BDV:=IL\cap BD. Since BDBD is the polar of LL with respect to ω\omega, and P′QP'Q is the diameter of ω\omega perpendicular to BDBD, the standard harmonic property of poles and polars gives that P′,V,Q,LP',V,Q,L form a harmonic range on line ILIL. Because inversion and projection from a point both preserve cross-ratio, this harmonic relation transfers to the pencils through JJ and through AA.