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Response from Дима[+++]
Answer: 2avc/a+v a+c c+c Let the original triangle be ABC, the bisectors bases are A1, B1, C1 By the property of the bisector A1C/A1B=A/c/c Therefore A1C/BC = c/ c+c Similarly B1C/AC = a/ a+c Therefore the ratio of area of triangle A1B1C to area of triangle ABC: S A1BC1C /S ABC = av/ a+c in+c= av a+v / a+v a+c in+c Similarly S ABC1C1 /S ABC = av c+v / a+v a+c in+c and S A1BC1 /S ABC = as a+c / a+v a+c in+c Note now, that S A1BC1C +S AB1C1 +S A1BC1 +S A1BC1=S ABC Therefore S A1BC1 /S ABC=1 S A1BC1 +S ABC1 +S A1BC1 /S ABC=1-av a+v / a+v a+c in+c -as a+v / a+v a+c in+c -as a+c / a+v a+c in+c=2av a+v a+c in+c?
Response from 0[+++++]
Answer: 2abc a+c a+c c+c Let the initial triangle -ABC bisect the base -A1 B1 C1 By the property of the bisector A1C A1B=Ac ABC ABC=c Therefore A1C BC=c in+c Similarly BC ABC=a a+c Therefore the ratio of area of triangle A1B1C to area of triangle ABC: S A1BC1C S ABC=a+c a+c c+c=a+c a+c a+c c+c Similarly S ABC1C1 S ABC=a+c a+c a+c a+c c+c and S A1BC1 S ABC=a+c a+c a+c a+c c+c Note now that S A1BC +S ABC1 +S ABC1 +S A1BC1 +S A1BC1=S ABC Therefore S A1BC1 S ABC=1 S A1BC +S ABC1 +S ABC1 +S A1BC1 S ABC=1- avs a+v a+v a+c v+c -vs s+v a+v a+c v+c -as a+v a+v a+c v+c=2 avs a+v a+v a+c v+c
Answer: 2abc a+c a+c c+c Let the initial triangle -ABC bisect the base -A1 B1 C1 By the property of the bisector A1C A1B=Ac ABC ABC=c Therefore A1C BC=c in+c Similarly BC ABC=a a+c Therefore the ratio of area of triangle A1B1C to area of triangle ABC: S A1BC1C S ABC=a+c a+c c+c=a+c a+c a+c c+c Similarly S ABC1C1 S ABC=a+c a+c a+c a+c c+c and S A1BC1 S ABC=a+c a+c a+c a+c c+c Note now that S A1BC +S ABC1 +S ABC1 +S A1BC1 +S A1BC1=S ABC Therefore S A1BC1 S ABC=1 S A1BC +S ABC1 +S ABC1 +S A1BC1 S ABC=1- avs a+v a+v a+c v+c -vs s+v a+v a+c v+c -as a+v a+v a+c v+c=2 avs a+v a+v a+c v+c