△ABC中,AB=2,AC=1,点D在BC上,若BD:DC=2:1,求AD的取值范围老师说至少十种解

网友回答

△ABC中,AB=2,AC=1,点D在BC上,BD:DC=2:1=AB:AC,
∴AD是∠BAC的平分线.
设AD=t,∠BAC=2a,
1)由S△ABC=S△BAD+S△CAD得
sin2a=(t/2)(2+1)sina,
∴t=(4/3)cosa,a∈(0,π/2),
∴AD=t的取值范围是(0,4/3).
2)由余弦定理,BD^2=4+t^2-4tcosa,CD^2=1+t^2-2tcosa,
BD=2CD,
∴4+t^2-4tcosa=4（1+t^2-2tcosa）,
整理得3t^2-4tcosa=0,t>0,∴t=(4/3)cosa.余者仿上.
3）由余弦定理,BD^2+t^2-2t*BDcosADB=4,①
CD^2+t^2-2t*CDcosADC=1,②
cosADB+cosADC=0,BD=2CD,
①+2*②,BD^2+2CD^2+3t^2=6,
∴t^2=2-2CD^2,
AB-AC=1