АС1/С1В=1/1, ВА1/А1С=3/7, АВ1/В1С=1/3, S A1B1C1=S ABC - S AC1B1 - S C1BA1 - S A1CB1, обе части уравнения делим на S ABC
S A1B1C1 / S ABC = 1 - (S AC1B1/S ABC) - (S C1BA1/ S ABC) - (S A1CB1/S ABC)
S ABC=1/2*AB*AC*sinA, S AB1C1=1/2*AC1*AB1*sinA, AB=AC1+C1B=1+1=2, AC=AB1+B1C=1+3=4, S AB1C1/S ABC=(AC1*AB1)/(AB*AC)=(1*1)/(2*4)=1/8,
S ABC=1/2*AB*BC*sinB, S C1BA1=1/2*C1B*BA1*sinB, BC=BA1+A1C=3+7=10,
S C1BA1/S ABC=(C1B*BA1)/(AB*BC)=(1*3)/(2*10)=3/20,
S ABC=1/2*AC*BC*sinC, S A1CB1=1/2*A1C*B1C*sinC, S A1CB/S ABC=(A1C*B1C) / (AC*BC)=(7*3)/(4*10)=21/40,
S A1B1C1/S ABC=1-1/8-3/20-21/40=8/40=1/5, или S ABC/S A1B1C1=5/1
В треугольнике ABC ∠C = 120°, CK—биссектриса.
Доказать, что 1 / CK = 1 / AC+1 / BC. || 1 / lc = 1 / a + 1 / b ||
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CK = 2*AC*BC*cos(∠ACB /2) / (AC+BC)
CK= 2*AC*BC*cos(120°/2) / (AC + BC) || cos60° =1 /2 ||
CK= AC*BC / (AC+BC) ⇔ 1 / CK = (AC+BC) / AC*BC
1 / CK = AC / AC*BC + BC / AC*BC
1 / CK = 1 / AC+ 1 / BC ч. т. д.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
* * * P.S. ∠ACB = ∠C ; ACK =∠BCK =∠ ACB /2 = ∠C /2
CK = Lc = 2abcos(∠C/2) / (a+b) * * *
действительно :
S(ΔACB) =S(ΔACK) + S(ΔBCK) ;
(1/2)*AC*BC*sin∠C=(1/2)*AC*CK*sin(∠C/2) + (1/2)*BC*CK*sin∠C/2)
(1/2)*AC*BC*sin∠C =(1/2)*CK*sin(∠C/2) *(AC + BC)
* * * ! sin2α = 2sinα*cosα * * *
* * * sin∠C = sin(2*∠C/2) = 2sin(∠C/2)*cos(∠C/2) * * *
2AC*BC*cos(∠C/2) = CK* (AC + BC) ;
CK =2AC*BC*cos(∠C/2) / (AC+BC) || Lc=2abcos(∠C/2)/(a+b) ||