求证：d垂+2?d外=3?d重.
分析：这里用三角法.设△ABC外接圆
半径为1,三个内角记为A,B,
C.易知d外=OO1+OO2+OO3
=cosA+cosB+cosC,
∴2d外=2(cosA+cosB+cosC).①
∵AH1=sinB?AB=sinB?(2sinC)=2sinB?sinC,
同样可得BH2?CH3.
∴3d重=△ABC三条高的和
=2?(sinB?sinC+sinC?sinA+sinA?sinB) ②
∴=2,
∴HH1=cosC?BH=2?cosB?cosC.
同样可得HH2,HH3.
∴d垂=HH1+HH2+HH3
=2(cosB?cosC+cosC?cosA+cosA?cosB) ③
欲证结论,观察①、②、③,
须证(cosB?cosC+cosC?cosA+cosA?cosB)+( cosA+ cosB+ cosC)=sinB?sinC+sinC?sinA+sinA?sinB.即可.