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Inner Mongolia mathematics middle school entrance examination paper
Solution: O is OD⊥AB at point D, OE⊥AC at point E, and OF⊥BC at point F.

∵OA, OB and OC are the three bisectors of △ABC.

∴OD=OE=OF,

∫△ABC's three sides AB, BC and CA are 40, 50 and 60 respectively.

∴s△abo:s△bco:s△cao=( 1/2ab? OD): (BC 1/2? OF):( 1/2 AC? OE)=AB:BC:AC=40:50:60=4:5:6

So the answer is: 4: 5: 6.