在三角形ABC中,abc分别为ABC的对边,
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A,B,C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa,b,c2bcosB=acosC\uff0bccosA\uff0c\u6b63\u5f26\u5b9a\u7406\uff0c\u52192sinBcosB=sinAcosC\uff0bsinCcosA=sinB\uff0ccosB=1/2\uff0cB=60\u00b0\u3002\u5ef6\u957f\u4e2d\u7ebfBD\u5230E\uff0c\u6784\u9020\u25a1ABCE\uff0c\u5219BE²\uff0bAC²=2(BA²\uff0bBC²)\uff0c\u90a3\u53ea\u8981\u6c42BC\u6700\u5927\uff0cb²=a²\uff0bc²\uff0dac=16\uff0d3ac\uff0cac\u2264[(a\uff0bc)/2]²
\u6839\u636e\u9898\u610f\uff0c\u4e24\u5411\u91cf\u5e73\u884c\uff0c\u5219\u6709\uff1a
[sin^2(B+C)/2]/(cos2A+7/2)=1/4
4cos^2(A/2\uff09=cos2A+7/2
2(1+cosA)=cos2A+7/2
cos2A-2cosA+3/2=0
2cos^2A-2cosA+1/2=0
4cos^2A-4cosA+1=0
(2cosA-1)^2=0
\u6240\u4ee5\uff1a
cosA=1/2.
\u6240\u4ee5A=60\u5ea6\u3002
a=\u221a3,b+c=3
\u56e0\u4e3a\uff1a
b+c=3
\u6240\u4ee5
b^2+2bc+c^2=9
\u6709\u4f59\u5f26\u5b9a\u7406\uff1acosA=1/2=(b^2+c^2-a^2)/2bc=(9-2bc-3)/2bc
\u6240\u4ee5bc=2
\u6240\u4ee5\u9762\u79ef=1/2*bc*sin60=\u221a3/2.
向量m(a-b,1)和向量n(b-c,1)平行
所以：a-b=b-c
所以：a+c=2b
S=acsinB/2=ac*(4/5)/2=3/2
所以：ac=15/4
所以：2b=a+c>=2√(ac)=√15
所以：b>=√15/2
由余弦定理得：b^2=a^2+c^2-2accosB=(a+c)^2-2ac(1+cosB)
所以；b^2=4b^2-(15/2)*(1+cosB)
整理得：b^2=5(1+cosB)/2>=(√15/2)^2=15/4
所以：cosB>1/2
因为：sinB=4/5
所以：cosB=3/5>1/2
所以：b^2=5(1+cosB)/2=5(1+3/5)/2=4
所以：b=2
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