设O为锐角△ABC的外心,R为△ABC的外接圆半径,AO,BO,CO的延长线分别交BC,CA,AB于点D,E,F,求证:(1/AD)+(1/BE)+(1/CF)=(2/R)
证明 设K,M,N分别是锐角△ABC边BC,CA,AB上的中点,连OK,OM,ON.
∵AO=BO=CO=R,OK=R*cosA,OM=R*cosB,ON=R*cosC.
∠DOK=｜(B-C)｜,∠EOM=｜(C-A)｜,∠FON=｜(A-B)｜.
∴AD=AO+OK/cos(B-C)=R+R*cosA/cos(B-C)
=2R*sinB*sinC/cos(B-C).
同样可得:
BE=2R*sinC*sinA/cos(C-A);
CF=2R*sinA*sinB/cos(A-B).
故 1/AD+1/BE+1/CF
=[1/(2R)]*[cos(B-C)/sinB*sinC+cos(C-A)/sinC*sinA+cos(A-B)/sinA*sinB]
=[1/(2R)]*[cotB*cotC+cotC*cotA+cotA*cotB+3]
=[1/((2R)]*(1+3)=2/R
最后一步用到恒等式:
cotB*cotC+cotC*cotA+cotA*cotB=1.
另一种证法
设ha,hb,hc分别是△ABC边BC,CA,AB上的高,BC=a,CA=b,AB=c,S表示其面积.
∵1/AD=cos(B-C)/ha,1/BE=cos(C-A)/hb,1/CF=cos(A-B)/hc
S=(sinA*bc)/2=2R^2*4sinA*sinB*sinC
∴1/AD+1/BE+1/CF
=a*cos(B-C)/(2S)+b*cos(C-A)/(2S)+c*cos(A-B)/(2S)
=[a*cos(B-C)+b*cos(C-A)+c*cos(A-B)]/(2S)
=R*[sinA*con(B-C)+sinB*cos(C-A)+sinC*cos(A-B)]/S
=(R/S)*[sin(B+C)*con(B-C)+sin(C+A)*cos(C-A)+sin(A+B)*cos(A-B)]
=(R/S)*[sin(2A)+sin(2B)+sin(2C)
=(R/S)*[4sinA*sinB*sinC]
=(R/S)*[S/(2R^2)]=2/R.