∠ A+ ∠ B+ ∠ C=180 °
∠ A= ∠ C
cos ∠ A=cos ∠C=sqrt(2/3)
cos^2 ∠ A=cos^2 ∠ C=2/3
sin^2 ∠ A=sin^2 ∠ C=1/3
cos ∠ B=cos(180 ° - ∠ A- ∠ B)=cos(180 ° -2 ∠ A)=
формула: 2 sin^2 α =1-cos2 α
=(1/2)*sin^2 ∠ A=1/6
Пусть АВ=BC=4x
Тогда CD=x
BD=3x
По теореме косинусов:
AD^2=AB^2+BD^2-2*AD*BD*cos ∠ B
(3/4)^2=(4x)^2+(3x)^2-2*4x*3x*(1/6)
(9/16)=21x^2
x^2=1/48
AB=4/sqrt(48)=1/sqrt(3)
cos ∠ B=1/6
sin ∠ B=sqrt(35)/6
S_( Δ ABC)=(1/2)*AB*BC*sin ∠ B)=(1/2)*(1/sqrt(3))*(1/sqrt(3))*(sqrt(35)/6)=...