1
p=AM=BM-BA=BC/2-CD - ( 1)
q=DN-DA=DC/2+BC=BC-CD/2 - (2)
(2)*2-( 1): 3BC/2=2q-p, which means BC=2(2q-p)/3.
(2)-( 1)*2: 3CD/2=q-2p, that is, CD=2(q-2p)/3.
2
Let the coordinates of the midpoint D on the BC side be (x, y, z), then: x=(8+0)/2=4, y=(0+8)/2=4, and z=(0+6)/2=3.
So the vector ad = (4 4,4,3)-(0,0,2) = (4,4, 1)
Therefore, the length of the BC side midline AD = | ad | = sqrt (16+16+1) = sqrt (33).
-
Or: AB = (8 8,0,0)-(0,0,2) = (8,0,2), AC = (0,8,6)-(0,0,2) = (0,8,4).
According to the parallelogram rule of vectors, the median line of BC side AD = (AB+AC)/2 = ((8,0,-2)+(0,8,4))/2.
= (4,4, 1), so | ad | = sqrt (16+1) = sqrt (33)