1、在△ABC中 ,∠A=π/6,D是BC边上任意一点（D与B,C不重合）,且向量AB的模^2=向量AD模^2 + 向量

我用~AB代表AB向量.
1.由已知有
AB² = AD²+~BD·~DC
= AD²+BD×DC
∵ AB² = AD²+BD²-2AD×BD×cos∠ADB
∴ AD²+BD×DC = AD²+BD²-2AD×BD×cos∠ADB
DC = BD - 2ADcos∠ADB
注意,如果∠B为钝角或直角,那么在图中易知 BD - ADcos∠ADB ≤ 0 ,那么 DC = BD - 2ADcos∠ADB ＜ 0 ,不可能,所以∠B必定为锐角,则有
DC = BD + 2ADcos∠ADC
BD + DC = 2BD + 2ADcos∠ADC
1/2BC = BD + ADcos∠ADC
注意,这意味着BC边的中线和垂线重合,有
AB = BC ,∠B = ∠C = 5π/12
2.(1) ∠A = ∠B ,则 ACcos∠A = 1/2AB = 1/2c
∵ AB·~AC = 1
∴ AB×AC×cos∠A = 1
1/2c² = 1
c = √2
(2) AB+~AC| = √6
AB+~AC|² = 6
AB+~AC)² = 6
AB²+~AC²+2~AB·~AC = 6
2 + b² + 2 = 6
b = √2
∴ a = b = c =√2
∴ S△ABC = 1/2×(√2)²sin60º
= 1/2√3