Let ABC be a triangle, I - its incenter, Ha, Hb, Hc the orthocenters of triangles BIC, CIA, AIB. Prove that AHb.BHc.CHa=AHc.BHa.CHb.
Proof:
AHb_|_CI_|_BHa, so <CAHb=<CBHa=90-<C/2 ( 1 ), similarly <ABHc=<ACHb ( 2 ) and
<BAHc=<BCHa ( 3 ). Applying sin theorem in triangle BCHa we get BHa/CHa=sin(<BCHa)/sin(<CBHa) ( 4 ), from triangle CAHb:
CHb/AHb=sin(<CAHb)/sin(<ACHb) ( 5 ) and from triangle ABHc:
AHc/BHc=sin(<ABHc)/sin(<BAHc) ( 6 ).
Multiplying side by side relations (4)-(6) and taking into account relations (1)-(3) we get the desired relation.
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