解
假設(shè)三角形的外接圓的直徑為d,對(duì)應(yīng)角的邊分別為a,b,c
則a/sinA=b/sinB=c/sinC=d
所以
sinA=a/d
sinB=b/d
sinC=c/d
因?yàn)?br/>sinA=sin(B+C)
sinC=sin(A+B)
所以
sin(B+C)/sin(A+B)=sin(A-B)/sin(B-C)
即sin(B+C)sin(B-C)=sin(A-B)sin(A+B)
(sinBcosC+cosBsinC)(sinBcosC-cosBsinC)=(sinAcosB-cosAsinB)(sinAcosB+cosAsinB)
sin^2Bcos^2C-cos^2Bsin^2C=sin^2Acos^2B-cos^2Asin^2B
將系數(shù)為負(fù)的移項(xiàng)
sin^2Bcos^2C+cos^2Asin^2B=sin^2Acos^2B+cos^2Bsin^2C
sin^2B(cos^2C+cos^2A)=cos^2B(sin^2A+sin^2C)
根據(jù)sin^2B+cos^2B=1
可以得到
cos^2B=1-sin^2B
所以
sin^2B(cos^2C+cos^2A)=(1-sin^2B)(sin^2A+sin^2C)
將上式拆分
sin^2B(cos^2C+cos^2A)=(sin^2A+sin^2C)-sin^2B(sin^2A+sin^2C)
將系數(shù)為負(fù)的移項(xiàng)
sin^2B(cos^2C+cos^2A+sin^2A+sin^2C)=sin^2A+sin^2C
因?yàn)?br/>cos^2A+sin^2A=1
cos^2C+sin^2C=1
所以
2sin^2B=sin^2A+sin^2C
所以
2(b/d)^2=(a/d)^2+(c/d)^2
即2b^2=a^2+c^2
所以命題得證