Question

在三角形中点D在直线CB的延长线上,且CD（向量）=4BD（向量）=rAB（向量）+sAC（向量）,则r

Answer 1

CD = 4BD=rAB+SAC

B,C,D is on the straight line

for k1+k2=1

AD = k1AB+k2AC

AB+BD = k1AB+k2AC

AB+ (1/4)CD =k1AB+k2AC

CD = [4(k1-1)]AB + (4k2)AC

r= 4(k1-1),s= 4k2

r+s = 4(k1-1)+ 4k2

= 4(k1+k2) -4

=4-4 =0