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The thing is I'm better with discrete problems than geometry, and so I rarely gets the solution.
I have checked the source again. It seems to actually be: (BC + CA)(BC^2 + CA^2 - AB^2) = 2*BC*CA^2. (First term is BC not BA).
Sorry for the fault!
The simple proof of (BC+CA)(BC^2+CA^2-AB^2)=2*BC*CA^2:
1) AC/CB=AL/LB - angle bisector theorem
2) AL/LB*BD/DC*CH/AH - Cevas theorem for ABC and AD, BH, CL; BD=DC => AL/LB=AH/HC
3) CH=BC*cosC - because BHC=90
4) {1), 2), 3)} =>AC/CB=AL/LB=AH/HC=AH/(CB*cosC) => AC*cosC=AH => (AH+HC)*cosC=AH => (1+CH/AH)cosC=1.
cosC=(AC^2+BC^2-AB^2)/(2*AC*BC) and CH/AH=CB/AC (by {1), 2)}) => (1+CB/AC)*(AC^2+BC^2-AB^2)/(2*AC*BC)=1 =>
(BC+CA)(BC^2+CA^2-AB^2) = 2*BC*CA^2. Done!
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