在三角形ABC中，AB=2AC,AD是∠A的平分线，且AD=kAC1求k的取值范围2若S三角形abc=1,问k取何值时，BC最短

做 CE⊥AD BH⊥AD分别交AD和AD延长线于点F、H.
设AC=a, CD=b, AD=c
1, ∠EAF=∠CAF CE⊥AD
AE=AC EF=CF 点E为AB中点
EF//BH, EF/BH=AE/AB, EF=BH/2
CF=EF, CF=BH/2
CF//BH, CD/BD=CF/BH, CD=BD/2, BD=2b
AB=2AC=2a
BD^2=AB^2+AD^2-2AB*AD*cos1
4b^2=4a^2+c^2-4ac*cos1 (1)
CD^2=AC^2+AD^2-2AC*AD*cos1
b^2=a^2+c^2-2ac*cos1 (2)
(2)*4-(1) 得 3c^2-4ac*cos1=0
c≠0 c=(4cos1/3)*a 即AD=(4cos1/3)*AC
k=4cos1/3 0 <∠BAC<180 0<∠1<90
故0 2, S(ABC)=AB*AC*sin(2∠1)/2
=2a^2sin1cos1 (k=4cos1/3 cos1=3k/4)
1=3ka^2sin1/2 a^2=2/3ksin1
BC^2=AB^2+AC^2-2AB*AC*cos(2∠1)
=4a^2+a^2-4a^2[1-2(sin1)^2]
=a^2+8a^2(sin1)^2 (a^2=2/3ksin1)
=2/3ksin1+16sin1/3k (k=4cos1/3)
=1/2sin1cos1+4sin1/cos1
=[1+8(sin1)^2]/(2sin1cos1)
=[(3sin1)^2+(cos1)^2]/(2sin1cos1)
≥6sin1cos1/2sin1cos1=3 (当3sin1=cos1时取最小值)
BC最小值为√3
3sin1=cos1 tan1=1/3 cos1=3√10/10
k=4cos1/3
=2√10/5

同问。。。