三边长分别是BC=17CA=18AB=19，过三角形内的点P作三边的垂线段PD、PE、PF，且BD+CE+AF=27,求BD+BF的值

Question

如题所述

Answer 1

用勾股定理
连接PC,PA,PB
PC^2=CE^2+PE^2=CD^2+PD^2 (1)
PA^2=PF^2+AF^2=AE^2+PE^2 (2)
PB^2=PD^2+BD^2=PF^2+BF^2 (3)
(1)+(2）+(3）得
BD^2+CE^2+AF^2=CD^2+BF^2+AE^2
=(17-BD)^2+(19-AF)^2+(18-CE)^2
34BD+36CE+38AF=17^2+18^2+19^2
34(BD+CE+AF)+2(CE+2AF)=974
34*27+2(CE+2AF)=974
CE+2AF=28
CE+AF=28-AF=28-(19-BF)=9+BF
27=BD+CE+AF=BD+9+BF
BD+BF=27-9=18